advanced_full.ipynb

{
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  {
   "cell_type": "markdown",
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   "source": [
    "Adapted from a workshop given at [POPL\n",
    "2008](https://www.cis.upenn.edu/~plclub/popl08-tutorial/).\n",
    "\n",
    "The NB language, back again\n",
    "===========================\n",
    "\n",
    "In this notebook, we will be working with a language very similar to the\n",
    "NB language from the first assignment.\n",
    "\n",
    "Definitions\n",
    "-----------\n",
    "\n",
    "### Grammar and terms\n",
    "\n",
    "The grammar of our language would be defined as follows:\n",
    "\n",
    "    t  ::= \"true\"                   terms\n",
    "         | \"false\"\n",
    "         | \"if\" t \"then\" t \"else\" t\n",
    "         | 0\n",
    "         | \"succ\" t\n",
    "         | \"pred\" t\n",
    "         | \"iszero\" t\n",
    "\n",
    "To represent the terms of this language in Coq, we will define `tm`, an\n",
    "`Inductive` data type:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Inductive tm : Set :=\n",
    "| tm_true : tm\n",
    "| tm_false : tm\n",
    "| tm_if : tm -> tm -> tm -> tm\n",
    "| tm_zero : tm\n",
    "| tm_succ : tm -> tm\n",
    "| tm_pred : tm -> tm\n",
    "| tm_iszero : tm -> tm."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Compare the two definitions. For every rule in the grammar, there is a\n",
    "corresponding *constructor*. Each constructor is a value in Coq that\n",
    "allows us to obtain values of type `tm`. In contrast to the previous\n",
    "notebook, here we annotate every constructor with a type. `tm_true` is a\n",
    "constructor corresponding to the terminal rule `\"true\"` - it takes no\n",
    "arguments and is therefore annotated with the simple type `tm`.\n",
    "`tm_succ` is a constructor corresponding to the `\"succ\" t` rule - since\n",
    "the rule has a single subterm, the constructor is a function from `tm`\n",
    "to `tm`.\n",
    "\n",
    "Using the above definition, we can create values corresponding to the\n",
    "terms in our language:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "(* Represents \"if (iszero 0) false true\" *)\n",
    "Check (tm_if (tm_iszero tm_zero) tm_false tm_true)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Definition of value\n",
    "\n",
    "Next, we want to define what it means to be a *value* in our language.\n",
    "While in the original NB language we did so through grammar rules, it’s\n",
    "equally valid to define a judgment which tells us which terms are\n",
    "boolean and numeric values (correspondingly, `bvalue` and `nvalue`):\n",
    "\n",
    "      ---------------  (b_true)\n",
    "      ⊢ bvalue (true)\n",
    "\n",
    "      ---------------- (b_false)\n",
    "      ⊢ bvalue (false)\n",
    "\n",
    "      \n",
    "      ---------- (n_zero)\n",
    "      ⊢ nvalue 0\n",
    "      \n",
    "         ⊢ nvalue t\n",
    "      ----------------- (n_succ)\n",
    "      ⊢ nvalue (succ t)\n",
    "\n",
    "Recall that from Curry-Howard correspondence we know that types\n",
    "correspond to propositions and values correspond to proofs. Therefore,\n",
    "we can represent the above judgements in Coq by defining types\n",
    "corresponding to the judgments. Those types are `bvalue t` and\n",
    "`nvalue t`. Being able to create a well-typed value of type `nvalue t`\n",
    "is the same as being able to construct a proof that a given term is an\n",
    "`nvalue`; same notion applies to `bvalue t`.\n",
    "\n",
    "We define said types as follows:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Inductive bvalue : tm -> Prop := \n",
    "| b_true : bvalue tm_true \n",
    "| b_false : bvalue tm_false.\n",
    "\n",
    "Inductive nvalue : tm -> Prop :=\n",
    "| n_zero : nvalue tm_zero\n",
    "| n_succ : forall t,\n",
    "    nvalue t ->\n",
    "    nvalue (tm_succ t)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Those definitions should look similar to the inference rules above,\n",
    "although they may also look slightly confusing. Before trying to\n",
    "understand their every part, it may help to see how they are meant to be\n",
    "used.\n",
    "\n",
    "Again, the *type* `nvalue t` represents the *proposition* that `t` is a\n",
    "numeric value. For instance, `nvalue (tm_succ tm_zero)` represents the\n",
    "proposition that the successor of zero (or simply one) is a numeric\n",
    "value. To show that this proposition is true, we need to construct a\n",
    "value of said type. We can do that as follows:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Check (n_succ tm_zero n_zero)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You should now go back to the definitions and try to understand how they\n",
    "represent their corresponding inference rules.\n",
    "\n",
    "One thing that may still be puzzling are the `Set` and `tm -> Prop`\n",
    "annotation. If you are suspecting that the second one mentions `tm ->`\n",
    "because the corresponding type is a *type-level function* from `tm` to\n",
    "`Prop`, you’d be correct:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Check nvalue.\n",
    "Check nvalue tm_zero."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The difference between `Set` and `Prop` is much more subtle and\n",
    "fundamental. To put it briefly, types annotated as `Set` are meant to be\n",
    "used as data types, while those annotated as `Prop` are meant to be used\n",
    "as propositions. Fully explaining this distinction is beyond the scope\n",
    "of this course, but the above intuition should serve you well enough.\n",
    "\n",
    "As the last thing in this section, we will (finally) define what it\n",
    "means to be a value. If you recall that `T \\/ S` is the data type\n",
    "corresponding to the proof that either `T` or `S`, the definition is\n",
    "simple enough:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Definition value (t : tm) : Prop :=\n",
    "  bvalue t \\/ nvalue t."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Operational semantics\n",
    "\n",
    "Having defined `tm`s and `value`s, we can define call-by-value\n",
    "operational semantics for our language. We will define an inductive data\n",
    "type `eval (t : tm) (t' : tm) : Prop` corresponding to the proposition\n",
    "that `t` evaluates to `t'` in a single step. The definition is as\n",
    "follows:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Inductive eval : tm -> tm -> Prop :=\n",
    "| e_iftrue : forall t2 t3,\n",
    "    eval (tm_if tm_true t2 t3) t2\n",
    "| e_iffalse : forall t2 t3,\n",
    "    eval (tm_if tm_false t2 t3) t3\n",
    "| e_if : forall t1 t1' t2 t3,\n",
    "    eval t1 t1' ->\n",
    "    eval (tm_if t1 t2 t3) (tm_if t1' t2 t3)\n",
    "| e_succ : forall t t',\n",
    "    eval t t' ->\n",
    "    eval (tm_succ t) (tm_succ t')\n",
    "| e_predzero :\n",
    "    eval (tm_pred tm_zero) tm_zero\n",
    "| e_predsucc : forall t,\n",
    "    nvalue t ->\n",
    "    eval (tm_pred (tm_succ t)) t\n",
    "| e_pred : forall t t',\n",
    "    eval t t' ->\n",
    "    eval (tm_pred t) (tm_pred t')\n",
    "| e_iszerozero :\n",
    "    eval (tm_iszero tm_zero) tm_true\n",
    "| e_iszerosucc : forall t,\n",
    "    nvalue t ->\n",
    "    eval (tm_iszero (tm_succ t)) tm_false\n",
    "| e_iszero : forall t t',\n",
    "    eval t t' ->\n",
    "    eval (tm_iszero t) (tm_iszero t')."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "If you don’t feel comfortable with Coq syntax yet, compare the above\n",
    "with the definition of beta-reduction from our first assignment.\n",
    "\n",
    "Next, we define the multi-step evaluation relation `eval_many`,\n",
    "corresponding to multi-step beta-reduction.\n",
    "\n",
    "Its inference rules are:\n",
    "\n",
    "     -------------  (m_refl)\n",
    "     eval_many t t \n",
    "     \n",
    "     eval t t'    eval_many t' u\n",
    "     ---------------------------  (m_step)\n",
    "           eval_many t u\n",
    "\n",
    "And its definition is:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Inductive eval_many : tm -> tm -> Prop :=\n",
    "| m_refl : forall t,\n",
    "    eval_many t t\n",
    "| m_step : forall t t' u,\n",
    "    eval t t' ->\n",
    "    eval_many t' u ->\n",
    "    eval_many t u."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We say that a term is a `normal_form` if there is no term to which it\n",
    "can step. We can define this in Coq as follows:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Definition normal_form (t : tm) : Prop :=\n",
    "  ~ exists t', eval t t'."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This definition uses negation `~` and an existential qualifier, notions\n",
    "which we will explain later.\n",
    "\n",
    "### Exercises\n",
    "\n",
    "**Exercise** Multi-step evaluation is often defined as the “reflexive,\n",
    "transitive closure” of single-step evaluation. Write an inductively\n",
    "defined relation `eval_rtc` that corresponds to that verbal description.\n",
    "\n",
    "In case you get stuck or need a hint, you can find solutions to all the\n",
    "exercises near the bottom of the file.\n",
    "\n",
    "**Exercise** Sometimes it is more convenient to use a big-step semantics\n",
    "for a language. Add the remaining constructors to finish the inductive\n",
    "definition `full_eval` for the big-step semantics that corresponds to\n",
    "the small-step semantics defined by `eval`. Build the inference rules so\n",
    "that `full_eval t v` logically implies both `eval_many t v` and\n",
    "`value v`. In order to do this, you may need to add the premise\n",
    "`nvalue v` to the appropriate cases.\n",
    "\n",
    "Hint: You should end up with a total of 8 cases.\n",
    "\n",
    "    Inductive full_eval : tm -> tm -> Prop :=\n",
    "    | f_value : forall v,\n",
    "        value v ->\n",
    "        full_eval v v\n",
    "    | f_iftrue : forall t1 t2 t3 v,\n",
    "        full_eval t1 tm_true ->\n",
    "        full_eval t2 v ->\n",
    "        full_eval (tm_if t1 t2 t3) v\n",
    "    | f_succ : forall t v,\n",
    "        nvalue v ->\n",
    "        full_eval t v ->\n",
    "        full_eval (tm_succ t) (tm_succ v).\n",
    "\n",
    "Proofs\n",
    "------\n",
    "\n",
    "So far, we’ve only seen proofs represented in Coq as\n",
    "manually-constructed values. For any non-trivial proof value, it’s\n",
    "rather inconvenient to manually construct it.\n",
    "\n",
    "Proof values are most easily built interactively, using tactics to\n",
    "manipulate a proof state. A proof state consists of a set of goals\n",
    "(propositions or types for which you must produce an inhabitant), each\n",
    "with a context of hypotheses (inhabitants of propositions or types you\n",
    "are allowed to use). A proof state begins initially with one goal (the\n",
    "statement of the lemma you are trying to prove) and no hypotheses. A\n",
    "goal can be solved, and thereby eliminated, when it exactly matches one\n",
    "of hypotheses in the context. A proof is completed when all goals are\n",
    "solved.\n",
    "\n",
    "Tactics can be used for forward reasoning (which, roughly speaking,\n",
    "means modifying the hypotheses of a context while leaving the goal\n",
    "unchanged) or backward reasoning (replacing the current goal with one or\n",
    "more new goals in simpler contexts). Given the level of detail required\n",
    "in a formal proof, it would be ridiculously impractical to complete a\n",
    "proof using forward reasoning alone. However it is usually both possible\n",
    "and practical to complete a proof using backward reasoning alone.\n",
    "Therefore, we focus almost exclusively on backward reasoning in this\n",
    "tutorial. Of course, most people naturally use a significant amount of\n",
    "forward reasoning in their thinking process, so it may take you a while\n",
    "to become accustomed to getting by without it.\n",
    "\n",
    "We use the keyword `Lemma` to state a new proposition we wish to prove.\n",
    "(`Theorem` and `Fact` are exact synonyms for `Lemma`.) The keyword\n",
    "`Proof`, immediately following the statement of the proposition,\n",
    "indicates the beginning of a proof script. A proof script is a sequence\n",
    "of tactic expressions, each concluding with a `.`. Once all of the goals\n",
    "are solved, we use the keyword `Qed` to record the completed proof. If\n",
    "the proof is incomplete, we may tell Coq to accept the lemma on faith by\n",
    "using `Admitted` instead of `Qed`.\n",
    "\n",
    "We now proceed to introduce the specific proof tactics.\n",
    "\n",
    "### Implication and universal quantification\n",
    "\n",
    "        - [intros]\n",
    "        - [apply]\n",
    "        - [apply with (x := ...)]\n",
    "\n",
    "Recall that both implication and universal quantification correspond to\n",
    "function types and values. Accordingly, we can use the `intros` tactic\n",
    "to move universally quantified variables and implication antecedents\n",
    "from the goal into the context as hypotheses.\n",
    "\n",
    "If our current goal corresponds to a conclusion of some implication `P`,\n",
    "we can use the `apply P` tactic to prove our goal by proving the\n",
    "antecedents of `P`. If you’d suspect from the name of the tactic that\n",
    "this corresponds to applying a function, you’d be correct. Using `apply`\n",
    "allows building a proof value from the bottom up.\n",
    "\n",
    "#### Example 1\n",
    "\n",
    "In the following example, we will create a value corresponding to a\n",
    "(still) simple proposition. Step through every cell below to see how\n",
    "this value is constructed."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma e_succ_pred_succ : forall t,\n",
    "  nvalue t ->\n",
    "  eval (tm_succ (tm_pred (tm_succ t))) (tm_succ t).\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** Let [t] be a [tm]. *)\n",
    "  intros t."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** Assume that [t] is an [nvalue] (and let's call that\n",
    "      assumption [Hn] for future reference). *)\n",
    "  intros Hn."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** By [e_succ], in order to prove our conclusion, it suffices\n",
    "      to prove that [eval (tm_pred (tm_succ t)) t]. *)\n",
    "  Check e_succ.\n",
    "  apply e_succ."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** That, in turn, can be shown by [e_predsucc], if we are\n",
    "      able to show that [nvalue t]. *)\n",
    "  Check e_predsucc.\n",
    "  apply e_predsucc."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** But, in fact, we assumed [nvalue t]. *)\n",
    "  apply Hn."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "At this point, we have successfully concluded our proof;\n",
    "`e_succ_pred_succ` is a value that can be used like any other value we\n",
    "have seen so far. This value corresponds to the following proof tree:\n",
    "\n",
    "                            nvalue t\n",
    "                 ---------------------------- (e_predsucc)\n",
    "                 eval (tm_pred (tm_succ t)) t\n",
    "        ------------------------------------------------ (e_succ)\n",
    "        eval (tm_succ (tm_pred (tm_succ t))) (tm_succ t)\n",
    "\n",
    "We can see the value we have constructed with the following command:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Print e_succ_pred_succ."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Compare the value to the proof script above. Observe how function\n",
    "application in the value corresponds to usages of `apply` tactic.\n",
    "\n",
    "#### Example 2\n",
    "\n",
    "Now consider, for a moment, the rule `m_step`:\n",
    "\n",
    "        eval t t'  eval_many t' u\n",
    "        ------------------------- (m_step)\n",
    "              eval_many t u\n",
    "\n",
    "If we have a goal such as `eval_many e1 e2`, we should be able to use\n",
    "`apply m_step` in order to replace it with the goals `eval e1 t'` and\n",
    "`eval_many t' e2`. But what exactly is `t'` here? When and how is it\n",
    "chosen? It stands to reason the conclusion is justified if we can come\n",
    "up with any `t'` for which the premises can be justified.\n",
    "\n",
    "Now we note that, in the Coq syntax for the type of `m_step`, all three\n",
    "variables `t`, `t'`, and `u` are universally quantified. The tactic\n",
    "`apply m_step` will use pattern matching between our goal and the\n",
    "conclusion of `m_step` to find the only possible instantiation of `t`\n",
    "and `u`. However, `apply m_step` will raise an error since it does not\n",
    "know how it should instantiate `t'`. In this case, the `apply` tactic\n",
    "takes a `with` clause that allows us to provide this instantiation. This\n",
    "is demonstrated in the proof below.\n",
    "\n",
    "Observe how this works in the proof script below. The proof tree here\n",
    "gives a visual representation of the proof term we are going to\n",
    "construct and the proof script has again been annotated with the steps\n",
    "in English.\n",
    "\n",
    "        Letting   s = tm_succ\n",
    "                  p = tm_pred\n",
    "                lem = e_succ_pred_succ,\n",
    "\n",
    "                nvalue t\n",
    "        - - - - - - - - - - - - (lem)    --------------------- (m_refl)\n",
    "        eval (s (p (s t))) (s t)         eval_many (s t) (s t)\n",
    "        ------------------------------------------------------ (m_step)\n",
    "                      eval_many (s (p (s t))) (s t)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_succ_pred_succ : forall t,\n",
    "  nvalue t ->\n",
    "  eval_many (tm_succ (tm_pred (tm_succ t))) (tm_succ t).\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** Let [t] be a [tm], and assume [nvalue t]. *)\n",
    "  intros t Hn."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** By [m_step], to show our conclusion, it suffices to find\n",
    "      some [t'] for which\n",
    "        [eval (tm_succ (tm_pred (tm_succ t))) t']\n",
    "      and\n",
    "        [eval t' (tm_succ t)].\n",
    "      Let us choose [t'] to be [tm_succ t]. *)\n",
    "  Check m_step.\n",
    "  apply m_step with (t' := tm_succ t)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    (** By the lemma [e_succ_pred_succ], to show\n",
    "          [eval (tm_succ (tm_pred (tm_succ t))) (tm_succ t)],\n",
    "        it suffices to show [nvalue t]. *)\n",
    "    Check e_succ_pred_succ.\n",
    "    apply e_succ_pred_succ."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    (** And, in fact, we assumed [nvalue t]. *)\n",
    "    apply Hn."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    (** Moreover, by the rule [m_refl], we also may conclude\n",
    "        [eval (tm_succ t) (tm_succ t)]. *)\n",
    "    Check m_refl.\n",
    "    apply m_refl."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Example 3\n",
    "\n",
    "Any tactic like `apply` that takes the name of a constructor or lemma as\n",
    "an argument can just as easily be given a more complicated expression as\n",
    "an argument. Thus, we may use function application to construct proof\n",
    "objects on the fly in these cases. Observe how this technique can be\n",
    "used to rewrite the proof of the previous lemma.\n",
    "\n",
    "Although, we have eliminated one use of `apply`, this is not necessarily\n",
    "an improvement over the previous proof. However, there are cases where\n",
    "this technique is quite valuable."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_succ_pred_succ_alt : forall t,\n",
    "  nvalue t ->\n",
    "  eval_many (tm_succ (tm_pred (tm_succ t))) (tm_succ t).\n",
    "Proof.\n",
    "  intros t Hn."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  Check m_step.\n",
    "  apply (m_step\n",
    "         (tm_succ (tm_pred (tm_succ t)))\n",
    "         (tm_succ t)\n",
    "         (tm_succ t)\n",
    "        )."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    Check e_succ_pred_succ.\n",
    "    apply (e_succ_pred_succ t Hn)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    apply t."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    Check m_refl.\n",
    "    apply (m_refl (tm_succ t))."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lab 1\n",
    "\n",
    "Write proof scripts for the following lemmas, following the plain\n",
    "language descriptions.\n",
    "\n",
    "These lemmas will be useful in later proofs."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_one : forall t1 t2,\n",
    "  eval t1 t2 ->\n",
    "  eval_many t1 t2."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** Let [t1] and [t2] be terms, and assume [eval t1 t2].  We\n",
    "      may conclude [eval_many t1 t2] by [m_step] if we can find\n",
    "      a term [t'] such that [eval t1 t'] and [eval_many t' t2].\n",
    "      We will choose [t'] to be [t2].  Now we can show\n",
    "      [eval t1 t2] by our assumption, and we can show\n",
    "      [eval_many t2 t2] by [m_refl]. *)\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_two : forall t1 t2 t3,\n",
    "  eval t1 t2 ->\n",
    "  eval t2 t3 ->\n",
    "  eval_many t1 t3."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** Let [t1], [t2], and [t3] be terms.  Assume [eval t1 t2]\n",
    "      and [eval t2 t3].  By [m_step], we may conclude that\n",
    "      [eval_many t1 t3] if we can find a term [t'] such that\n",
    "      [eval t1 t'] and [eval_many t' t3].  Let's choose [t'] to\n",
    "      be [t2].  We know [eval t1 t2] holds by assumption.  In\n",
    "      the other case, by the lemma [m_one], to show [eval_many\n",
    "      t2 t3], it suffices to show [eval t2 t3], which is one of\n",
    "      our assumptions.  *)\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_iftrue_step : forall t t1 t2 u,\n",
    "  eval t tm_true ->\n",
    "  eval_many t1 u ->\n",
    "  eval_many (tm_if t t1 t2) u."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** Let [t], [t1], [t2], and [u] be terms.  Assume that\n",
    "      [eval t tm_true] and [eval_many t1 u].  To show\n",
    "      [eval_many (tm_if t t1 t2) u], by [m_step], it suffices to\n",
    "      find a [t'] for which [eval (tm_if t t1 t2) t'] and\n",
    "      [eval_many t' u].  Let us choose [t'] to be\n",
    "      [tm_if tm_true t1 t2].  Now we can use [e_if] to show that\n",
    "      [eval (tm_if t t1 t2) (tm_if tm_true t1 t2)] if we can\n",
    "      show [eval t tm_true], which is actually one of our\n",
    "      assumptions.  Moreover, using [m_step] once more, we can\n",
    "      show [eval_many (tm_if tm_true t1 t2) u] where [t'] is\n",
    "      chosen to be [t1].  Doing so leaves us to show\n",
    "      [eval (tm_if tm_true t1 t2) t1] and [eval_many t1 u].  The\n",
    "      former holds by [e_iftrue] and the latter holds by\n",
    "      assumption. *)\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Working with Definitions\n",
    "\n",
    "        - [unfold]\n",
    "\n",
    "There is a notion of equivalence on Coq terms that arises from the\n",
    "conversion rules of the underlying calculus of constructions. It is\n",
    "sometimes useful to be able to replace one term in a proof with an\n",
    "equivalent one. For instance, we may want to replace a defined name with\n",
    "its definition. This sort of replacement can be done the tactic\n",
    "`unfold`. This tactic can be used to manipulate the goal or the\n",
    "hypotheses."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Definition strongly_diverges t :=\n",
    "  forall u, eval_many t u -> ~ normal_form u."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma unfold_example : forall t t',\n",
    "  strongly_diverges t ->\n",
    "  eval t t' ->\n",
    "  strongly_diverges t'.\n",
    "Proof.\n",
    "  intros t t' Hd He.\n",
    "   unfold strongly_diverges. intros u Hm.\n",
    "   unfold strongly_diverges in Hd.\n",
    "   apply Hd. apply m_step with (t' := t').\n",
    "    apply He.\n",
    "    apply Hm.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** In reality, many tactics will perform conversion\n",
    "automatically as necessary. Try removing the uses of `unfold` from the\n",
    "above proof to check which ones were necessary.\n",
    "\n",
    "### Working with Conjunction and Disjunction\n",
    "\n",
    "        - [split]\n",
    "        - [left]\n",
    "        - [right]\n",
    "        - [destruct] (for conjunction and disjunction)\n",
    "\n",
    "**Example** If `H` is the name of a conjunctive hypothesis, then\n",
    "`destruct H as p` will replace the hypothesis `H` with its components\n",
    "using the names in the pattern `p`. Observe the pattern in the example\n",
    "below."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_two_conj : forall t t' t'',\n",
    "  eval t t' /\\ eval t' t'' ->\n",
    "  eval_many t t''.\n",
    "Proof.\n",
    "  intros t t' t'' H.\n",
    "   destruct H as [ He1 He2 ].\n",
    "   apply m_two with (t2 := t').\n",
    "    apply He1.\n",
    "    apply He2.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Patterns may be nested to break apart nested structures.\n",
    "Note that infix conjunction is right-associative, which is significant\n",
    "when trying to write nested patterns. We will later see how to use\n",
    "`destruct` on many different sorts of hypotheses."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_three_conj : forall t t' t'' t''',\n",
    "  eval t t' /\\ eval t' t'' /\\ eval t'' t''' ->\n",
    "  eval_many t t'''.\n",
    "Proof.\n",
    "  intros t t' t'' t''' H.\n",
    "   destruct H as [ He1 [ He2 He3 ] ].\n",
    "   apply m_step with (t' := t').\n",
    "    apply He1.\n",
    "    apply m_two with (t2 := t'').\n",
    "      apply He2.\n",
    "      apply He3.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** If your goal is a conjunction, use `split` to break it apart\n",
    "into two separate subgoals."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_three : forall t t' t'' t''',\n",
    "  eval t t' ->\n",
    "  eval t' t'' ->\n",
    "  eval t'' t''' ->\n",
    "  eval_many t t'''.\n",
    "Proof.\n",
    "  intros t t' t'' t''' He1 He2 He3.\n",
    "   apply m_three_conj with (t' := t') (t'' := t'').\n",
    "    split.\n",
    "      apply He1.\n",
    "      split.\n",
    "        apply He2.\n",
    "        apply He3.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Hint: You might find lemma `m_three` useful here."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_if_iszero_conj : forall v t2 t2' t3 t3',\n",
    "  nvalue v /\\ eval t2 t2' /\\ eval t3 t3' ->\n",
    "  eval_many (tm_if (tm_iszero tm_zero) t2 t3) t2' /\\\n",
    "  eval_many (tm_if (tm_iszero (tm_succ v)) t2 t3) t3'.\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** If the goal is a disjunction, we can use the `left` or\n",
    "`right` tactics to solve it by proving the left or right side of the\n",
    "conclusion."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma true_and_succ_zero_values :\n",
    "  value tm_true /\\ value (tm_succ tm_zero).\n",
    "Proof.\n",
    "  unfold value. split.\n",
    "    left. apply b_true.\n",
    "    right. apply n_succ. apply n_zero.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** If we have a disjunction in the context, we can use\n",
    "`destruct` to reason by cases on the hypothesis. Note the syntax of the\n",
    "associated pattern."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma e_if_true_or_false : forall t1 t2,\n",
    "  eval t1 tm_true \\/ eval t1 tm_false ->\n",
    "  eval_many (tm_if t1 t2 t2) t2.\n",
    "Proof.\n",
    "  intros t1 t2 H. destruct H as [ He1 | He2 ].\n",
    "    apply m_two with (t2 := tm_if tm_true t2 t2).\n",
    "      apply e_if. apply He1.\n",
    "      apply e_iftrue.\n",
    "    apply m_two with (t2 := tm_if tm_false t2 t2).\n",
    "      apply e_if. apply He2.\n",
    "      apply e_iffalse.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lab 2\n",
    "\n",
    "Work on the following exercise.\n",
    "\n",
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma two_values : forall t u,\n",
    "  value t /\\ value u ->\n",
    "    bvalue t \\/\n",
    "    bvalue u \\/\n",
    "    (nvalue t /\\ nvalue u)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** We know [value t] and [value u], which means either\n",
    "      [bvalue t] or [nvalue t], and either [bvalue u] or\n",
    "      [nvalue u].  Consider the case in which\n",
    "      [bvalue t] holds. Then one of the disjuncts of our\n",
    "      conclusion is proved.  Next, consider the case in which\n",
    "      [nvalue t] holds.  Now consider the subcase where\n",
    "      [bvalue u] holds. ... *)\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** `destruct` can be used on propositions with implications.\n",
    "This will have the effect of performing `destruct` on the conclusion of\n",
    "the implication, while leaving the hypotheses of the implication as\n",
    "additional subgoals."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma destruct_example : forall bv t t' t'',\n",
    "  bvalue bv ->\n",
    "  (value bv -> eval t t' /\\ eval t' t'') ->\n",
    "  eval_many t t''.\n",
    "Proof.\n",
    "  intros bv t t' t'' Hbv H. destruct H as [ H1 H2 ].\n",
    "    Show 2.\n",
    "    unfold value. left. apply Hbv.\n",
    "    apply m_two with (t2 := t').\n",
    "      apply H1.\n",
    "      apply H2.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Tip** After applying a tactic that introduces multiple subgoals, it is\n",
    "sometimes useful to see not only the subgoals themselves but also their\n",
    "hypotheses. Adding the command `Show n.` to your proof script to cause\n",
    "Coq to display the nth subgoal in full.\n",
    "\n",
    "Reasoning by Cases and Induction\n",
    "--------------------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    - [destruct] (for inductively defined propositions)\n",
    "    - [induction]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Use `destruct` to reason by cases on an inductively defined\n",
    "datatype or proposition.\n",
    "\n",
    "Note: It is possible to supply `destruct` with a pattern in these\n",
    "instances also. However, the patterns become increasingly complex for\n",
    "bigger inductive definitions; so it is often more practical to omit the\n",
    "pattern (thereby letting Coq choose the names of the terms and\n",
    "hypotheses in each case), in spite of the fact that this adds an element\n",
    "of fragility to the proof script (since the proof script will mention\n",
    "names that were system-generated)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma e_iszero_nvalue : forall v,\n",
    "  nvalue v ->\n",
    "  eval (tm_iszero v) tm_true \\/\n",
    "  eval (tm_iszero v) tm_false.\n",
    "Proof.\n",
    "  intros v Hn."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  destruct Hn."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* Case [n_zero].\n",
    "     Note how [v] becomes [tm_zero] in the goal. *)\n",
    "    left."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    apply e_iszerozero."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* Case [n_succ].\n",
    "     Note how [v] becomes [tm_succ v] in the goal. *)\n",
    "    right."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    apply e_iszerosucc. apply Hn.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** You can use `induction` to reason by induction on an\n",
    "inductively defined datatype or proposition. This is the same as\n",
    "`destruct`, except that it also introduces an induction hypothesis in\n",
    "the inductive cases."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_iszero : forall t u,\n",
    "  eval_many t u ->\n",
    "  eval_many (tm_iszero t) (tm_iszero u).\n",
    "Proof.\n",
    "  intros t u Hm. induction Hm.\n",
    "    apply m_refl.\n",
    "    apply m_step with (t' := tm_iszero t').\n",
    "      apply e_iszero. apply H.\n",
    "      apply IHHm.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lab 3\n",
    "\n",
    "Work on the following exercise.\n",
    "\n",
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_trans : forall t t' u,\n",
    "  eval_many t t' ->\n",
    "  eval_many t' u ->\n",
    "  eval_many t u."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (** We proceed by induction on the derivation of\n",
    "      [eval_many t t'].\n",
    "      Case [m_refl]: Since [t] and [t'] must be the same, our\n",
    "        conclusion holds by assumption.\n",
    "      Case [m_step]: Now let's rename the [t'] from the lemma\n",
    "        statement to [u0] (as Coq likely will) and observe that\n",
    "        there must be some [t'] (from above the line of the\n",
    "        [m_step] rule) such that [eval t t'] and\n",
    "        [eval_many t' u0].  Our conclusion follows from from\n",
    "        an application of [m_step] with our new [t'] and our\n",
    "        induction hypothesis, which allows us to piece together\n",
    "        [eval_many t' u0] and [eval_many u0 u] to get\n",
    "        [eval_many t' u]. *)\n",
    "Proof."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** It is possible to use `destruct` not just on hypotheses but\n",
    "on any lemma we have proved. If we have a lemma\n",
    "\n",
    "          lemma1 : P /\\ Q\n",
    "\n",
    "then we can use the tactic\n",
    "\n",
    "          destruct lemma1 as [ H1 H2 ].\n",
    "\n",
    "to continue our proof with `H1 : P` and `H2 : Q` in our context. This\n",
    "works even if the lemma has antecedents (they become new subgoals);\n",
    "however it fail if the lemma has a universal quantifier, such as this:\n",
    "\n",
    "          lemma2 : forall x, P(x) /\\ Q(x)\n",
    "\n",
    "However, remember that we can build a proof of `P(e) /\\ Q(e)` (which can\n",
    "be destructed) using the Coq expression `lemma2 e`. So we need to phrase\n",
    "our tactic as\n",
    "\n",
    "          destruct (lemma2 e) as [ H1 H2 ].\n",
    "\n",
    "An example of this technique is below."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_iszero_nvalue : forall t v,\n",
    "  nvalue v ->\n",
    "  eval_many t v ->\n",
    "  eval_many (tm_iszero t) tm_true \\/\n",
    "  eval_many (tm_iszero t) tm_false.\n",
    "Proof.\n",
    "  intros t v Hnv Hm.\n",
    "   destruct (e_iszero_nvalue v) as [ H1 | H2 ].\n",
    "    apply Hnv.\n",
    "    left. apply m_trans with (t' := tm_iszero v).\n",
    "      apply m_iszero. apply Hm.\n",
    "      apply m_one. apply H1.\n",
    "    right. apply m_trans with (t' := tm_iszero v).\n",
    "      apply m_iszero. apply Hm.\n",
    "      apply m_one. apply H2.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Prove the following lemma.\n",
    "\n",
    "Hint: You may be interested in some previously proved lemmas, such as\n",
    "`m_one` and `m_trans`.\n",
    "\n",
    "Note: Even though this lemma is in a comment, its solution is also at\n",
    "the bottom. (Coq will give an error if we leave it uncommented since it\n",
    "mentions the `eval_rtc` relation, which was the solution to another\n",
    "exercise.)\n",
    "\n",
    "    Lemma eval_rtc_many : forall t u,\n",
    "      eval_rtc t u ->\n",
    "      eval_many t u.\n",
    "\n",
    "**Exercise** Prove the following lemma.\n",
    "\n",
    "    Lemma eval_many_rtc : forall t u,\n",
    "      eval_many t u ->\n",
    "      eval_rtc t u.\n",
    "\n",
    "**Exercise** Prove the following lemma.\n",
    "\n",
    "    Lemma full_eval_to_value : forall t v,\n",
    "      full_eval t v ->\n",
    "      value v.\n",
    "\n",
    "Working with Existential Quantification\n",
    "---------------------------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    - [exists]\n",
    "    - [destruct] (for existential propositions)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Use `exists` to give the witness for an existential\n",
    "quantifier in your goal."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma if_bvalue : forall t1 t2 t3,\n",
    "  bvalue t1 ->\n",
    "  exists u, eval (tm_if t1 t2 t3) u.\n",
    "Proof.\n",
    "  intros t1 t2 t3 Hb. destruct Hb.\n",
    "    exists t2. apply e_iftrue.\n",
    "    exists t3. apply e_iffalse.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** You may use `destruct` to break open an existential\n",
    "hypothesis."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_two_exists : forall t u,\n",
    "  (exists w, eval t w /\\ eval w u) ->\n",
    "  eval_many t u.\n",
    "Proof.\n",
    "  intros t u H.\n",
    "   destruct H as [ w He ].\n",
    "   destruct He as [ He1 He2 ].\n",
    "   apply m_two with (t2 := w).\n",
    "    apply He1.\n",
    "    apply He2.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Tip: We can combine patterns that destruct existentials with\n",
    "patterns that destruct other logical connectives.\n",
    "\n",
    "Here is the same proof with just one use of `destruct`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_two_exists' : forall t u,\n",
    "  (exists w, eval t w /\\ eval w u) ->\n",
    "  eval_many t u.\n",
    "Proof.\n",
    "  intros t u H. destruct H as [ w [ He1 He2 ] ].\n",
    "   apply m_two with (t2 := w).\n",
    "    apply He1.\n",
    "    apply He2.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Tip: We give patterns to the `intros` tactic to destruct\n",
    "hypotheses as we introduce them.\n",
    "\n",
    "Here is the same proof again without any uses of `destruct`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_two_exists'' : forall t u,\n",
    "  (exists w, eval t w /\\ eval w u) ->\n",
    "  eval_many t u.\n",
    "Proof.\n",
    "  intros t u [ w [ He1 He2 ] ].\n",
    "   apply m_two with (t2 := w).\n",
    "    apply He1.\n",
    "    apply He2.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma value_can_expand : forall v,\n",
    "  value v ->\n",
    "  exists u, eval u v.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lab 4\n",
    "\n",
    "    Work on the following exercise.\n",
    "\n",
    "**Exercise** Tip: You should find the lemma `m_iszero` useful. Use\n",
    "`Check m_iszero.` if you’ve forgotten its statement."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma exists_iszero_nvalue : forall t,\n",
    "  (exists nv, nvalue nv /\\ eval_many t nv) ->\n",
    "  exists bv, eval_many (tm_iszero t) bv.\n",
    "Proof.\n",
    "  (** There exists some [nv] such that [nvalue nv].  Consider\n",
    "      the case where [nv] is [tm_zero].  Then choose [bv] to\n",
    "      be [tm_true].  By [m_trans], we can show that\n",
    "      [eval_many (tm_iszero t) tm_true] by showing\n",
    "      [eval_many (tm_iszero t) (tm_iszero tm_zero)] and\n",
    "      [eval_many (tm_iszero tm_zero) tm_true].  The former\n",
    "      follows from [m_iszero] and our assumption.  The latter\n",
    "      follows from [m_one] and the rule [e_iszerozero].  On the\n",
    "      other hand, in the case where [nv] is built from\n",
    "      [tm_succ], we choose [bv] to be [tm_false] and the proof\n",
    "      follows similarly. *)\n",
    "\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Working with Negation\n",
    "---------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    - [unfold not]\n",
    "    - [destruct] (for negation)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** The standard library defines an uninhabited type `False` and\n",
    "defines `not P` to stand for `P -> False`. Furthermore, Coq defines the\n",
    "notation `~ P` to stand for `not P`. (Such notations only affect parsing\n",
    "and printing – Coq views `not P` and `~ P` as being syntactically\n",
    "equal.)\n",
    "\n",
    "The most basic way to work with negated statements is to unfold `not`\n",
    "and treat `False` just as any other proposition.\n",
    "\n",
    "(Note how multiple definitions can be unfolded with one use of `unfold`.\n",
    "Also, as noted earlier, many uses of `unfold` are not strictly\n",
    "necessary. You can try deleting the uses from the proof below to check\n",
    "that the proof script still works.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma normal_form_succ : forall t,\n",
    "  normal_form (tm_succ t) ->\n",
    "  normal_form t.\n",
    "Proof.\n",
    "  intros t Hnf.\n",
    "   unfold normal_form. unfold not.\n",
    "   unfold normal_form, not in Hnf.\n",
    "   intros [ t' H' ]. apply Hnf.\n",
    "   exists (tm_succ t'). apply e_succ. apply H'.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma normal_form_to_forall : forall t,\n",
    "  normal_form t ->\n",
    "  forall u, ~ eval t u.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma normal_form_from_forall : forall t,\n",
    "  (forall u, ~ eval t u) ->\n",
    "  normal_form t.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** If you happen to have `False` as a hypothesis, you may use\n",
    "`destruct` on that hypothesis to solve your goal."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma False_hypothesis : forall v,\n",
    "  False ->\n",
    "  value v.\n",
    "Proof.\n",
    "  intros v H. destruct H.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Recalling that `destruct` can be used on propositions with\n",
    "antecedents and that negation is simply an abbreviation for an\n",
    "implication, using `destruct` on a negated hypothesis has the derived\n",
    "behavior of replacing our goal with the proposition that was negated in\n",
    "our context.\n",
    "\n",
    "Tip: We actually don’t even need to do the unfolding below because\n",
    "`destruct` would have done it for us."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma destruct_negation_example : forall t v,\n",
    "  value v ->\n",
    "  eval t tm_zero ->\n",
    "  (value v -> normal_form t) ->\n",
    "  eval tm_true tm_false.\n",
    "Proof.\n",
    "  intros t v Hnv He Hnf.\n",
    "   unfold normal_form, not in Hnf.\n",
    "  (* As usual, unfolding was optional here. *)\n",
    "   destruct Hnf.\n",
    "    apply Hnv.\n",
    "    exists tm_zero. apply He.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** This one may be a bit tricky. Start by using `destruct` on\n",
    "one of your hypotheses."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma negation_exercise : forall v1 v2,\n",
    "  ~ (value v1 \\/ value v2) ->\n",
    "  ~ (~ bvalue v1 /\\ ~ bvalue v2) ->\n",
    "  eval tm_true tm_false.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Working with Equality\n",
    "---------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    - [reflexivity]\n",
    "    - [subst]\n",
    "    - [rewrite]\n",
    "    - [inversion] (on equalities)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** If you have an equality in your context, there are several\n",
    "ways to substitute one side of the equality for the other in your goal\n",
    "or in other hypotheses.\n",
    "\n",
    "If one side of the equality is a variable `x`, then the tactic `subst x`\n",
    "will replace all occurrences of `x` in the context and goal with the\n",
    "other side of the quality and will remove `x` from your context.\n",
    "\n",
    "Use `reflexivity` to solve a goal of the form `e = e`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma equality_example_1 : forall t1 t2 t3 u1 u2,\n",
    "  t1 = tm_iszero u1 ->\n",
    "  t2 = tm_succ u2 ->\n",
    "  t3 = tm_succ t2 ->\n",
    "  tm_if t1 t2 t3 =\n",
    "    tm_if (tm_iszero u1) (tm_succ u2) (tm_succ (tm_succ u2)).\n",
    "Proof.\n",
    "  intros t1 t2 t3 u1 u2 Heq1 Heq2 Heq3.\n",
    "   subst t1. subst t2. subst t3. reflexivity.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** If neither side of the equality in your context is a\n",
    "variable (or if you don’t want to discard the hypothesis), you can use\n",
    "the `rewrite` tactic to perform a substitution. The arrow after\n",
    "`rewrite` indicates the direction of the substitution. As demonstrated,\n",
    "you may perform rewriting in the goal or in a hypothesis."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma equality_example_2a : forall t u v,\n",
    "  tm_succ t = tm_succ u ->\n",
    "  eval (tm_succ u) v ->\n",
    "  eval (tm_succ t) v.\n",
    "Proof.\n",
    "  intros t u v Heq He. rewrite -> Heq. apply He.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma equality_example_2b : forall t u v,\n",
    "  tm_succ t = tm_succ u ->\n",
    "  eval (tm_succ u) v ->\n",
    "  eval (tm_succ t) v.\n",
    "Proof.\n",
    "  intros t u v Heq He. rewrite <- Heq in He. apply He.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** We also note that, analogously with `destruct`, we may use\n",
    "`rewrite` even with a hypothesis (or lemma) that has antecedents."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma equality_example_2c : forall t u v,\n",
    "  nvalue v ->\n",
    "  (nvalue v -> tm_succ t = tm_succ u) ->\n",
    "  eval (tm_succ u) v ->\n",
    "  eval (tm_succ t) v.\n",
    "Proof.\n",
    "  intros t u v Hnv Heq He. rewrite <- Heq in He.\n",
    "    apply He.\n",
    "    apply Hnv.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** If you need to derive additional equalities implied by an\n",
    "equality in your context (e.g., by the principle of constructor\n",
    "injectivity), you may use `inversion`. `inversion` is a powerful tactic\n",
    "that uses unification to introduce more equalities into your context.\n",
    "(You will observe that it also performs some substitutions in your\n",
    "goal.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma equality_example_3 : forall t u,\n",
    "  tm_succ t = tm_succ u ->\n",
    "  t = u.\n",
    "Proof.\n",
    "  intros t u Heq. inversion Heq. reflexivity.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma equality_exercise : forall t1 t2 t3 u1 u2 u3 u4,\n",
    "  tm_if t1 t2 t3 = tm_if u1 u2 u2 ->\n",
    "  tm_if t1 t2 t3 = tm_if u3 u3 u4 ->\n",
    "  t1 = u4.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** `inversion` will also solve a goal when unification fails on\n",
    "a hypothesis. (Internally, Coq can construct a proof of `False` from\n",
    "contradictory equalities.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma equality_example_4 :\n",
    "  tm_zero = tm_true ->\n",
    "  eval tm_true tm_false.\n",
    "Proof.\n",
    "  intros Heq. inversion Heq.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lab 5\n",
    "\n",
    "Work on `equality_exercise` above and `succ_not_circular` below.\n",
    "\n",
    "**Exercise** Note: `e1 <> e2` is a notation for `~ e1 = e2`, i.e., the\n",
    "two are treated as syntactically equal.\n",
    "\n",
    "Note: This is fairly trivial to prove if we have a size function on\n",
    "terms and some automation. With just the tools we have described so far,\n",
    "it requires just a little bit of work.\n",
    "\n",
    "Hint: The proof requires induction on `t`. (This is the first example of\n",
    "induction on datatypes, but it is even more straightforward than\n",
    "induction on propositions.) In each case, unfold the negation, pull the\n",
    "equality into the context, and use `inversion` to eliminate\n",
    "contradictory equalities."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma succ_not_circular : forall t,\n",
    "  t <> tm_succ t.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Reasoning by Inversion\n",
    "----------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    - [inversion] (on propositions)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** The `inversion` tactic also allows you to reason by\n",
    "inversion on an inductively defined proposition as in paper proofs: we\n",
    "try to match some proposition with the conclusion of each inference rule\n",
    "and only consider the cases (possibly none) where there is a successful\n",
    "unification. In those cases, we may use the premises of the inference\n",
    "rule in our reasoning.\n",
    "\n",
    "Since `inversion` may generate many equalities between variables, it is\n",
    "useful to know that using `subst` without an argument will perform all\n",
    "possible substitutions for variables. It is a little difficult to\n",
    "predict which variables will be eliminated and which will be kept by\n",
    "this tactic, but this is a typical sort of trade-off when using powerful\n",
    "tactics.\n",
    "\n",
    "(The use of `subst` in this proof is superfluous, but you can observe\n",
    "that it simplifies the context.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma value_succ_nvalue : forall t,\n",
    "  value (tm_succ t) ->\n",
    "  nvalue t.\n",
    "Proof.\n",
    "  intros t H. unfold value in H. destruct H as [ H1 | H2 ].\n",
    "  (* No unification is possible -- [inversion] solves goal. *)\n",
    "    inversion H1.\n",
    "  (* Just the [n_succ] cases unifies with H2. *)\n",
    "    inversion H2. subst. apply H0.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lab 6\n",
    "\n",
    "Work on the exercise below.\n",
    "\n",
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma inversion_exercise : forall t,\n",
    "  normal_form t ->\n",
    "  eval_many (tm_pred t) tm_zero ->\n",
    "  nvalue t.\n",
    "Proof.\n",
    "  (** By inversion on the [eval_many] relation, then conclusion\n",
    "      [eval_many (tm_pred t) tm_zero] must have been derived by\n",
    "      the rule [m_step], which means there is some [t'] for\n",
    "      which [eval (tm_pred t) t'] and [eval_many t' tm_zero].\n",
    "      Now, by inversion on the [eval] relation, there are only\n",
    "      three ways that [eval (tm_pred t) t'] could have been\n",
    "      derived:\n",
    "      * By [e_predzero], with [t] and [t'] both being equal to\n",
    "        [tm_zero].  Our conclusion follows from [n_zero].\n",
    "      * By [e_predsucc], with [t] being [tm_succ t0] where we\n",
    "        have [nvalue t0].  In this case, our conclusion is\n",
    "        provable with [n_succ].\n",
    "      * By [e_pred], with [t] taking an evaluation step.  This\n",
    "        contradicts our assumption that [t] is a normal form\n",
    "        (which can be shown by using [destruct] on that\n",
    "        assumption). *)\n",
    "\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Tip: Nested patterns will be useful here."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma contradictory_equalities_exercise :\n",
    "  (exists t, exists u, exists v,\n",
    "    value t /\\\n",
    "    t = tm_succ u /\\\n",
    "    u = tm_pred v) ->\n",
    "  eval tm_true tm_false.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma eval_fact_exercise : forall t1 t2,\n",
    "  eval (tm_iszero (tm_pred t1)) t2 ->\n",
    "  eval t2 tm_false ->\n",
    "  exists u, t1 = tm_succ u.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma normal_form_if : forall t1 t2 t3,\n",
    "  normal_form (tm_if t1 t2 t3) ->\n",
    "  t1 <> tm_true /\\ t1 <> tm_false /\\ normal_form t1.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Additional Important Tactics\n",
    "----------------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    - [generalize dependent]\n",
    "    - [assert]\n",
    "    - [;]\n",
    "    - [clear]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Sometimes we need to have a tactic that moves hypotheses\n",
    "from our context back into our goal. Often this is because we want to\n",
    "perform induction in the middle of a proof and will not get a\n",
    "sufficiently general induction hypothesis without a goal of the correct\n",
    "form. (To be specific, if we need to have an induction hypothesis with a\n",
    "`forall` quantifier in front, then we must make sure our goal has a\n",
    "`forall` quantifier in front at the time we invoke the `induction`\n",
    "tactic.) Observe how `generalize dependent` achieves this in the proof\n",
    "below, moving the variable `t` and all dependent hypotheses back into\n",
    "the goal. You may want to remove the use of `generalize dependent` to\n",
    "convince yourself that it is performing an essential role here."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma value_is_normal_form : forall v,\n",
    "  value v ->\n",
    "  normal_form v.\n",
    "Proof.\n",
    "  intros v [ Hb | Hn ] [ t He ].\n",
    "    destruct Hb.\n",
    "      inversion He.\n",
    "      inversion He.\n",
    "    generalize dependent t. induction Hn.\n",
    "      intros t He. inversion He.\n",
    "      intros u He. inversion He. subst. destruct (IHHn t').\n",
    "       apply H0.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Coq has many operations (called “tacticals”) to combine\n",
    "smaller tactics into larger ones.\n",
    "\n",
    "If `t1` and `t2` are tactics, then `t1; t2` is a tactic that executes\n",
    "`t1`, and then executes `t2` on subgoals left by or newly generated by\n",
    "`t1`. This can help to eliminate repetitious use of tactics. Two\n",
    "idiomatic uses are performing `subst` after `inversion` and performing\n",
    "`intros` after `induction`. More opportunities to use this tactical can\n",
    "usually be discovered after writing a proof. (It is worth noting that\n",
    "some uses of this tactical can make proofs less readable or more\n",
    "difficult to maintain. Alternatively, some uses can make proofs more\n",
    "readable or easier to maintain. It is always good to think about your\n",
    "priorities when writing a proof script.)\n",
    "\n",
    "Revise the proof for `value_is_normal_form` to include uses of the `;`\n",
    "tactical.\n",
    "\n",
    "**Example** Sometimes it is helpful to be able to use forward reasoning\n",
    "in a proof. One form of forward reasoning can be done with the tactic\n",
    "`assert`. `assert` adds a new hypothesis to the context but asks us to\n",
    "first justify it."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma nvalue_is_normal_form : forall v,\n",
    "  nvalue v ->\n",
    "  normal_form v.\n",
    "Proof.\n",
    "  intros v Hnv.\n",
    "   assert (value v) as Hv. right. apply Hnv.\n",
    "   apply value_is_normal_form. apply Hv.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** `assert` can also be supplied with a tactic that proves the\n",
    "assertion. We rewrite the above proof using this form."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma nvalue_is_normal_form' : forall v,\n",
    "  nvalue v ->\n",
    "  normal_form v.\n",
    "Proof.\n",
    "  intros v Hnv.\n",
    "   assert (value v) as Hv by (right; apply Hnv).\n",
    "   apply value_is_normal_form. apply Hv.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** The proof below introduces two new, simple tactics. First,\n",
    "the tactic `replace e1 with e2` performs a substitution in the goal and\n",
    "then requires that you prove `e2 = e1` as a new subgoal. This often\n",
    "allows us to avoid more cumbersome forms of forward reasoning. Second,\n",
    "the `clear` tactic discards a hypothesis from the context. Of course,\n",
    "this tactic is never needed, but it can be nice to use when there are\n",
    "complicated, irrelevant hypotheses in the context."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma single_step_to_multi_step_determinacy :\n",
    "  (forall t u1 u2, eval t u1 -> eval t u2 -> u1 = u2) ->\n",
    "  forall t v1 v2,\n",
    "    eval_many t v1 -> normal_form v1 ->\n",
    "    eval_many t v2 -> normal_form v2 ->\n",
    "    v1 = v2.\n",
    "Proof.\n",
    "  intros H t v1 v2 Hm1 Hnf1 Hm2 Hnf2. induction Hm1.\n",
    "    clear H. destruct Hm2.\n",
    "      reflexivity.\n",
    "      destruct Hnf1. exists t'. apply H.\n",
    "    destruct Hm2.\n",
    "      destruct Hnf2. exists t'. apply H0.\n",
    "      apply IHHm1; clear IHHm1.\n",
    "        apply Hnf1.\n",
    "        replace t' with t'0.\n",
    "          apply Hm2.\n",
    "          apply H with (t := t).\n",
    "            apply H1.\n",
    "            apply H0.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** This proof is lengthy and thus somewhat challenging. All of\n",
    "the techniques from this section will be useful; some will be essential.\n",
    "In particular, you will need to use `generalize dependent` at the\n",
    "beginning of the proof. You will find `assert` helpful in the cases\n",
    "where your assumptions are contradictory but none of them are in a\n",
    "negative form. In that situation, you can assert a negative statement\n",
    "that follows from your hypotheses (recall that `normal_form` is a\n",
    "negative statement). Finally, you will want to use the above lemma\n",
    "`nvalue_is_normal_form`. Good luck!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Theorem eval_deterministic : forall t t' t'',\n",
    "  eval t t' ->\n",
    "  eval t t'' ->\n",
    "  t' = t''.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Prove the following lemmas. The last is quite long, and you\n",
    "may wish to wait until you know more about automation.\n",
    "\n",
    "    Lemma full_eval_from_value : forall v w,\n",
    "      value v ->\n",
    "      full_eval v w ->\n",
    "      v = w.\n",
    "\n",
    "    Lemma eval_full_eval : forall t t' v,\n",
    "      eval t t' ->\n",
    "      full_eval t' v ->\n",
    "      full_eval t v.\n",
    "\n",
    "    Lemma full_eval_complete : forall t v,\n",
    "      value v ->\n",
    "      eval_many t v ->\n",
    "      full_eval t v.\n",
    "\n",
    "Basic Automation\n",
    "----------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "  - [eapply], [esplit]\n",
    "  - [auto], [eauto]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** You can use `eapply e` instead of `apply e with (x := e1)`.\n",
    "This will generate subgoals containing unification variables that will\n",
    "get unified during subsequent uses of `apply`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_if : forall t1 u1 t2 t3,\n",
    "  eval_many t1 u1 ->\n",
    "  eval_many (tm_if t1 t2 t3) (tm_if u1 t2 t3).\n",
    "Proof.\n",
    "  intros t1 u1 t2 t3 Hm. induction Hm.\n",
    "    apply m_refl.\n",
    "    eapply m_step.\n",
    "      apply e_if. apply H.\n",
    "      apply IHHm.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** You can use `esplit` to turn an existentially quantified\n",
    "variable in your goal into a unification variable."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma exists_pred_zero :\n",
    "  exists u, eval (tm_pred tm_zero) u.\n",
    "Proof.\n",
    "  esplit. apply e_predzero.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** The `auto` tactic solves goals that are solvable by any\n",
    "combination of\n",
    "\n",
    "-   `intros`\n",
    "-   `apply` (used on some local hypothesis)\n",
    "-   `split`, `left`, `right`\n",
    "-   `reflexivity`\n",
    "\n",
    "If `auto` cannot solve the goal, it will leave the proof state\n",
    "completely unchanged (without generating any errors).\n",
    "\n",
    "The lemma below is a proposition that has been contrived for the sake of\n",
    "demonstrating the scope of the `auto` tactic and does not say anything\n",
    "of practical interest. So instead of thinking about what it means, you\n",
    "should think about the operations that `auto` had to perform to solve\n",
    "the goal.\n",
    "\n",
    "Note: It is important to remember that `auto` does not destruct\n",
    "hypotheses! There are more advanced forms of automation available that\n",
    "do destruct hypotheses in some specific ways."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma auto_example : forall t t' t'',\n",
    "  eval t t' ->\n",
    "  eval t' t'' ->\n",
    "  (forall u, eval t t' -> eval t' u -> eval_many t u) ->\n",
    "  eval t' t \\/ t = t /\\ eval_many t t''.\n",
    "Proof.\n",
    "  auto.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** The `eauto` tactic solves goals that are solvable by some\n",
    "combination of\n",
    "\n",
    "-   `intros`\n",
    "-   `eapply` (used on some local hypothesis)\n",
    "-   `split`, `left`, `right`\n",
    "-   `esplit`\n",
    "-   `reflexivity`\n",
    "\n",
    "lemma has two significantly differences from the previous one, both of\n",
    "which render `auto` useless."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma eauto_example : forall t t' t'',\n",
    "  eval t t' ->\n",
    "  eval t' t'' ->\n",
    "  (forall u, eval t u -> eval u t'' -> eval_many t t'') ->\n",
    "  eval t' t \\/ (exists u, t = u) /\\ eval_many t t''.\n",
    "Proof.\n",
    "  eauto.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** You can enhance `auto` (or `eauto`) by appending\n",
    "`using x_1, ..., x_n`, where each `x_i` is the name of some constructor\n",
    "or lemma. Then `auto` will attempt to apply those constructors or lemmas\n",
    "in addition to the assumptions in the local context."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma eauto_using_example : forall t t' t'',\n",
    "  eval t t' ->\n",
    "  eval t' t'' ->\n",
    "  eval t' t \\/ t = t /\\ eval_many t t''.\n",
    "Proof.\n",
    "  eauto using m_step, m_one.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lab 7\n",
    "\n",
    "**Exercise** Go back and rewrite your proofs for `m_one`, `m_two`, and\n",
    "`m_iftrue_step`. You should be able to make them very succinct given\n",
    "what you know now.\n",
    "\n",
    "**Exercise** See how short you can make these proofs.\n",
    "\n",
    "Note: This is an exercise. We are not making the claim that shorter\n",
    "proofs are necessarily better!\n",
    "\n",
    "Hint: Remember that we can connect tactics in sequence with `;`.\n",
    "However, as you can imagine, figuring out the best thing to write after\n",
    "a `;` usually involves some trial and error."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma pred_not_circular : forall t,\n",
    "  t <> tm_pred t.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_succ : forall t u,\n",
    "  eval_many t u ->\n",
    "  eval_many (tm_succ t) (tm_succ u).\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma m_pred : forall t u,\n",
    "  eval_many t u ->\n",
    "  eval_many (tm_pred t) (tm_pred u).\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Go back and rewrite your proofs for `m_trans` and\n",
    "`two_values`. Pulling together several tricks you’ve learned, you should\n",
    "be able to prove `two_values` in one (short) line. Since this is a\n",
    "notebook, the easiest way for you to do that may be to copy the cells\n",
    "here.\n",
    "\n",
    "**Note** Sometimes there are lemmas or constructors that are so\n",
    "frequently needed by `auto` that we don’t want to have to add them to\n",
    "our `using` clause each time. Coq allows us to request that certain\n",
    "propositions that always be considered by `auto` and `eauto`.\n",
    "\n",
    "The following command adds four lemmas to the default search procedure\n",
    "of `auto`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Hint Resolve m_if m_succ m_pred m_iszero."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Constructors of inductively defined propositions are some of the most\n",
    "frequently needed by `auto`. Instead of writing\n",
    "\n",
    "        Hint Resolve b_true b_false.\n",
    "\n",
    "we may simply write\n",
    "\n",
    "        Hint Constructors bvalue.\n",
    "\n",
    "Let’s add all our constructors to `auto`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Hint Constructors bvalue nvalue eval eval_many."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "By default `auto` will never try to unfold definitions to see if a lemma\n",
    "or constructor can be applied. With the `Hint Unfold` command, we can\n",
    "instruct `auto` to try unfold definitions in the goal as it is working."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Hint Unfold value normal_form."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "There are a few more variants on the `Hint` command that can be used to\n",
    "further customize `auto`. You can learn about them in the Coq reference\n",
    "manual.\n",
    "\n",
    "Functions and Conversion\n",
    "------------------------"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "    - [Fixpoint/struct]\n",
    "    - [match ... end]\n",
    "    - [if ... then ... else ...]\n",
    "    - [simpl]\n",
    "    - [remember]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "In this section we start to use Coq as a programming language and learn\n",
    "how to reason about programs defined within Coq.\n",
    "\n",
    "**Example** Coq defines many datatypes in its standard libraries. Have a\n",
    "quick look now through the library `Datatypes` to see some of the basic\n",
    "ones, in particular `bool` and `nat`. (Note that constructors of the\n",
    "datatype `nat` are the letter `O` and the letter `S`. However, Coq will\n",
    "parse and print `nat`s using a standard decimal representation.)\n",
    "\n",
    "We define two more datatypes here that will be useful later."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Inductive bool_option : Set :=\n",
    "| some_bool : bool -> bool_option\n",
    "| no_bool : bool_option."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Inductive nat_option : Set :=\n",
    "| some_nat : nat -> nat_option\n",
    "| no_nat : nat_option."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** We can define simple (non-recursive) functions from one\n",
    "datatype to another using the `Definition` keyword. The `match`\n",
    "construct allows us to do case analysis on a datatype. The `match`\n",
    "expression has a first-match semantics and allows nested patterns;\n",
    "however, Coq’s type checker demands that pattern-matching be exhaustive.\n",
    "\n",
    "We define functions below for converting between Coq `bool`s and boolean\n",
    "values in our object language."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Definition tm_to_bool (t : tm) : bool_option :=\n",
    "  match t with\n",
    "  | tm_true => some_bool true\n",
    "  | tm_false => some_bool false\n",
    "  | _ => no_bool\n",
    "  end."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Definition bool_to_tm (b : bool) : tm :=\n",
    "  match b with\n",
    "  | true => tm_true\n",
    "  | false => tm_false\n",
    "  end."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Coq also has an `if/then/else` expression. It can be used,\n",
    "not just with the type `bool` but, in fact, with any datatype having\n",
    "exactly two constructors (the first constructor corresponding to the\n",
    "`then` branch and the second to the `else` branch). Thus, we can define\n",
    "a function `is_bool` as below."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Definition is_bool (t : tm) : bool :=\n",
    "  if tm_to_bool t then true else false."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** To define a recursive function, use `Fixpoint` instead of\n",
    "`Definition`.\n",
    "\n",
    "The type system will only allow us to write functions that terminate.\n",
    "The annotation `{struct t}` here informs the type-checker that\n",
    "termination is guaranteed because the function is being defined by\n",
    "structural recursion on `t`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Fixpoint tm_to_nat (t : tm) {struct t} : nat_option :=\n",
    "  match t with\n",
    "  | tm_zero => some_nat O\n",
    "  | tm_succ t1 =>\n",
    "      match tm_to_nat t1 with\n",
    "      | some_nat n => some_nat (S n)\n",
    "      | no_nat => no_nat\n",
    "      end\n",
    "  | _ => no_nat\n",
    "  end.\n",
    "\n",
    "Fixpoint nat_to_tm (n : nat) {struct n} : tm :=\n",
    "  match n with\n",
    "  | O => tm_zero\n",
    "  | S m => tm_succ (nat_to_tm m)\n",
    "  end."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Write a function `interp : tm -> tm` that returns the\n",
    "normal form of its argument according to the small-step semantics given\n",
    "by `eval`.\n",
    "\n",
    "Hint: You will want to use `tm_to_nat` (or another auxiliary function)\n",
    "to prevent stuck terms from stepping in the cases `e_predsucc` and\n",
    "`e_iszerosucc`.\n",
    "\n",
    "**Example** The tactic `simpl` (recursively) reduces the application of\n",
    "a function defined by pattern-matching to an argument with a constructor\n",
    "at its head. You can supply `simpl` with a particular expression if you\n",
    "want to prevent it from simplifying elsewhere."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma bool_tm_bool : forall b,\n",
    "  tm_to_bool (bool_to_tm b) = some_bool b.\n",
    "Proof.\n",
    "  intros b. destruct b.\n",
    "    simpl (bool_to_tm true). simpl. reflexivity.\n",
    "    (* It turns out that [simpl] is unnecessary above, since\n",
    "       [reflexivity] can automatically check that two terms are\n",
    "       convertible. *)\n",
    "    reflexivity.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** We can also apply the tactic `simpl` in our hypotheses."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma tm_bool_tm :forall t b,\n",
    "  tm_to_bool t = some_bool b ->\n",
    "  bool_to_tm b = t.\n",
    "Proof.\n",
    "  intros t b Heq. destruct t.\n",
    "    simpl in Heq. inversion Heq.\n",
    "     simpl. reflexivity.\n",
    "    (* As with [reflexivity], [inversion] can automatically\n",
    "       perform reduction on terms as necessary, so the above use\n",
    "       of [simpl] was optional. *)\n",
    "    inversion Heq. reflexivity.\n",
    "    simpl in Heq. inversion Heq.\n",
    "    (* Again, the above use of [simpl] was optional. *)\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma tm_to_bool_dom_includes_bvalue : forall bv,\n",
    "  bvalue bv -> exists b, tm_to_bool bv = some_bool b.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma tm_to_bool_dom_only_bvalue : forall bv b,\n",
    "  tm_to_bool bv = some_bool b -> bvalue bv.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Not all uses of `simpl` are optional. Sometimes they are\n",
    "necessary so that we can use the `rewrite` tactic. Observe, also, how\n",
    "using `rewrite` can automatically trigger a reduction if it creates a\n",
    "redex."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma nat_tm_nat : forall n,\n",
    "  tm_to_nat (nat_to_tm n) = some_nat n.\n",
    "Proof.\n",
    "  intros n. induction n.\n",
    "    reflexivity.\n",
    "    simpl. rewrite -> IHn. reflexivity.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Here’s an example where it is necessary to use `simpl` on a\n",
    "hypothesis. To trigger a reduction of a `match` expression in a\n",
    "hypothesis, we use the `destruct` tactic on the expression being\n",
    "matched."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma tm_nat_tm : forall t n,\n",
    "  tm_to_nat t = some_nat n ->\n",
    "  nat_to_tm n = t.\n",
    "Proof.\n",
    "  intros t. induction t; intros n Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq. reflexivity.\n",
    "    simpl in Heq. destruct (tm_to_nat t).\n",
    "      inversion Heq. simpl. rewrite -> IHt.\n",
    "        (* Note how we may use [rewrite] even on an equation\n",
    "           that is preceded by some other hypotheses. *)\n",
    "        reflexivity.\n",
    "        reflexivity.\n",
    "      inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma tm_to_nat_dom_includes_nvalue : forall v,\n",
    "  nvalue v -> exists n, tm_to_nat v = some_nat n.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma tm_to_nat_dom_only_nvalue : forall v n,\n",
    "  tm_to_nat v = some_nat n -> nvalue v.\n",
    "Proof.\n",
    "  (* to finish *)\n",
    "Admitted."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Example** Using the tactic `destruct` (or `induction`) on a complex\n",
    "expression (i.e., one that is not simply a variable) may not leave you\n",
    "with enough information for you to finish the proof. The tactic\n",
    "`remember` can help in these cases. Its usage is demonstrated below. If\n",
    "you are curious, try to finish the proof without `remember` to see what\n",
    "goes wrong."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Lemma remember_example : forall v,\n",
    "  eval_many\n",
    "    (tm_pred (tm_succ v))\n",
    "    (match tm_to_nat v with\n",
    "      | some_nat _ => v\n",
    "      | no_nat => tm_pred (tm_succ v)\n",
    "     end).\n",
    "Proof.\n",
    "  intros v. remember (tm_to_nat v) as x. destruct x.\n",
    "    apply m_one. apply e_predsucc.\n",
    "     eapply tm_to_nat_dom_only_nvalue.\n",
    "     rewrite <- Heqx. reflexivity.\n",
    "    apply m_refl.\n",
    "Qed."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Exercise** Prove the following lemmas involving the function `interp`\n",
    "from a previous exercise:\n",
    "\n",
    "    Lemma interp_reduces : forall t,\n",
    "      eval_many t (interp t).\n",
    "\n",
    "    Lemma interp_fully_reduces : forall t,\n",
    "      normal_form (interp t).\n",
    "\n",
    "Solutions to Exercises\n",
    "======================"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Inductive eval_rtc : tm -> tm -> Prop :=\n",
    "| r_eval : forall t t',\n",
    "    eval t t' ->\n",
    "    eval_rtc t t'\n",
    "| r_refl : forall t,\n",
    "    eval_rtc t t\n",
    "| r_trans : forall t u v,\n",
    "    eval_rtc t u ->\n",
    "    eval_rtc u v ->\n",
    "    eval_rtc t v.\n",
    "\n",
    "Inductive full_eval : tm -> tm -> Prop :=\n",
    "| f_value : forall v,\n",
    "    value v ->\n",
    "    full_eval v v\n",
    "| f_iftrue : forall t1 t2 t3 v,\n",
    "    full_eval t1 tm_true ->\n",
    "    full_eval t2 v ->\n",
    "    full_eval (tm_if t1 t2 t3) v\n",
    "| f_iffalse : forall t1 t2 t3 v,\n",
    "    full_eval t1 tm_false ->\n",
    "    full_eval t3 v ->\n",
    "    full_eval (tm_if t1 t2 t3) v\n",
    "| f_succ : forall t v,\n",
    "    nvalue v ->\n",
    "    full_eval t v ->\n",
    "    full_eval (tm_succ t) (tm_succ v)\n",
    "| f_predzero : forall t,\n",
    "    full_eval t tm_zero ->\n",
    "    full_eval (tm_pred t) tm_zero\n",
    "| f_predsucc : forall t v,\n",
    "    nvalue v ->\n",
    "    full_eval t (tm_succ v) ->\n",
    "    full_eval (tm_pred t) v\n",
    "| f_iszerozero : forall t,\n",
    "    full_eval t tm_zero ->\n",
    "    full_eval (tm_iszero t) tm_true\n",
    "| f_iszerosucc : forall t v,\n",
    "    nvalue v ->\n",
    "    full_eval t (tm_succ v) ->\n",
    "    full_eval (tm_iszero t) tm_false.\n",
    "\n",
    "Lemma m_one_sol : forall t t',\n",
    "  eval t t' ->\n",
    "  eval_many t t'.\n",
    "Proof.\n",
    "  intros t t' He. apply m_step with (t' := t').\n",
    "    apply He.\n",
    "    apply m_refl.\n",
    "Qed.\n",
    "\n",
    "Lemma m_two_sol : forall t t' t'',\n",
    "  eval t t' ->\n",
    "  eval t' t'' ->\n",
    "  eval_many t t''.\n",
    "Proof.\n",
    "  intros t t' t'' He1 He2. apply m_step with (t' := t').\n",
    "    apply He1.\n",
    "    apply m_one. apply He2.\n",
    "Qed.\n",
    "\n",
    "Lemma m_iftrue_step_sol : forall t t1 t2 u,\n",
    "  eval t tm_true ->\n",
    "  eval_many t1 u ->\n",
    "  eval_many (tm_if t t1 t2) u.\n",
    "Proof.\n",
    "  intros t t1 t2 u He Hm.\n",
    "   apply m_step with (t' := tm_if tm_true t1 t2).\n",
    "    apply e_if. apply He.\n",
    "    apply m_step with (t' := t1).\n",
    "      apply e_iftrue.\n",
    "      apply Hm.\n",
    "Qed.\n",
    "\n",
    "Lemma m_if_iszero_conj_sol : forall v t2 t2' t3 t3',\n",
    "  nvalue v /\\ eval t2 t2' /\\ eval t3 t3' ->\n",
    "  eval_many (tm_if (tm_iszero tm_zero) t2 t3) t2' /\\\n",
    "  eval_many (tm_if (tm_iszero (tm_succ v)) t2 t3) t3'.\n",
    "Proof.\n",
    "  intros v t2 t2' t3 t3' H.\n",
    "   destruct H as [ Hn [ He1 He2 ] ]. split.\n",
    "    apply m_three with\n",
    "     (t' := tm_if tm_true t2 t3) (t'' := t2).\n",
    "      apply e_if. apply e_iszerozero.\n",
    "      apply e_iftrue.\n",
    "      apply He1.\n",
    "    apply m_three with\n",
    "     (t' := tm_if tm_false t2 t3) (t'' := t3).\n",
    "      apply e_if. apply e_iszerosucc. apply Hn.\n",
    "      apply e_iffalse.\n",
    "      apply He2.\n",
    "Qed.\n",
    "\n",
    "Lemma two_values_sol : forall t u,\n",
    "  value t /\\ value u ->\n",
    "    bvalue t \\/\n",
    "    bvalue u \\/\n",
    "    (nvalue t /\\ nvalue u).\n",
    "Proof.\n",
    "  unfold value. intros t u H.\n",
    "   destruct H as [ [ Hb1 | Hn1 ] H2 ].\n",
    "    left. apply Hb1.\n",
    "    destruct H2 as [ Hb2 | Hn2 ].\n",
    "      right. left. apply Hb2.\n",
    "      right. right. split.\n",
    "        apply Hn1.\n",
    "        apply Hn2.\n",
    "Qed.\n",
    "\n",
    "Lemma m_trans_sol : forall t u v,\n",
    "  eval_many t u ->\n",
    "  eval_many u v ->\n",
    "  eval_many t v.\n",
    "Proof.\n",
    "  intros t u v Hm1 Hm2. induction Hm1.\n",
    "    apply Hm2.\n",
    "    apply m_step with (t' := t').\n",
    "      apply H.\n",
    "      apply IHHm1. apply Hm2.\n",
    "Qed.\n",
    "\n",
    "Lemma eval_rtc_many_sol : forall t u,\n",
    "  eval_rtc t u ->\n",
    "  eval_many t u.\n",
    "Proof.\n",
    "  intros t u Hr. induction Hr.\n",
    "    apply m_one. apply H.\n",
    "    apply m_refl.\n",
    "    apply m_trans with (t' := u).\n",
    "      apply IHHr1.\n",
    "      apply IHHr2.\n",
    "Qed.\n",
    "\n",
    "Lemma eval_many_rtc_sol : forall t u,\n",
    "  eval_many t u ->\n",
    "  eval_rtc t u.\n",
    "Proof.\n",
    "  intros t u Hm. induction Hm.\n",
    "    apply r_refl.\n",
    "    apply r_trans with (u := t').\n",
    "      apply r_eval. apply H.\n",
    "      apply IHHm.\n",
    "Qed.\n",
    "\n",
    "Lemma full_eval_to_value_sol : forall t v,\n",
    "  full_eval t v ->\n",
    "  value v.\n",
    "Proof.\n",
    "  intros t v Hf. induction Hf.\n",
    "    apply H.\n",
    "    apply IHHf2.\n",
    "    apply IHHf2.\n",
    "    right. apply n_succ. apply H.\n",
    "    right. apply n_zero.\n",
    "    right. apply H.\n",
    "    left. apply b_true.\n",
    "    left. apply b_false.\n",
    "Qed.\n",
    "\n",
    "Lemma value_can_expand_sol : forall v,\n",
    "  value v ->\n",
    "  exists u, eval u v.\n",
    "Proof.\n",
    "  intros v Hv. exists (tm_if tm_true v v). apply e_iftrue.\n",
    "Qed.\n",
    "\n",
    "Lemma exists_iszero_nvalue_sol : forall t,\n",
    "  (exists nv, nvalue nv /\\ eval_many t nv) ->\n",
    "  exists bv, eval_many (tm_iszero t) bv.\n",
    "Proof.\n",
    "  intros t [ nv [ Hnv Hm ]]. destruct Hnv.\n",
    "    exists tm_true.\n",
    "     apply m_trans with (t' := tm_iszero tm_zero).\n",
    "      apply m_iszero. apply Hm.\n",
    "      apply m_one. apply e_iszerozero.\n",
    "    exists tm_false.\n",
    "     apply m_trans with (t' := tm_iszero (tm_succ t0)).\n",
    "      apply m_iszero. apply Hm.\n",
    "      apply m_one. apply e_iszerosucc. apply Hnv.\n",
    "Qed.\n",
    "\n",
    "Lemma normal_form_to_forall_sol : forall t,\n",
    "  normal_form t ->\n",
    "  forall u, ~ eval t u.\n",
    "Proof.\n",
    "  unfold normal_form, not. intros t H u He.\n",
    "   apply H. exists u. apply He.\n",
    "Qed.\n",
    "\n",
    "Lemma normal_form_from_forall_sol : forall t,\n",
    "  (forall u, ~ eval t u) ->\n",
    "  normal_form t.\n",
    "Proof.\n",
    "  unfold normal_form, not. intros t H [ t' Het' ].\n",
    "   apply H with (u := t'). apply Het'.\n",
    "Qed. \n",
    "\n",
    "Lemma negation_exercise_sol : forall v1 v2,\n",
    "  ~ (value v1 \\/ value v2) ->\n",
    "  ~ (~ bvalue v1 /\\ ~ bvalue v2) ->\n",
    "  eval tm_true tm_false.\n",
    "Proof.\n",
    "  intros v1 v2 H1 H2. destruct H2.\n",
    "    split.\n",
    "      intros Hb. destruct H1. left. left. apply Hb.\n",
    "      intros Hb. destruct H1. right. left. apply Hb.\n",
    "Qed.\n",
    "\n",
    "Lemma equality_exercise_sol : forall t1 t2 t3 u1 u2 u3 u4,\n",
    "  tm_if t1 t2 t3 = tm_if u1 u2 u2 ->\n",
    "  tm_if t1 t2 t3 = tm_if u3 u3 u4 ->\n",
    "  t1 = u4.\n",
    "Proof.\n",
    "  intros t1 t2 t3 u1 u2 u3 u4 Heq1 Heq2.\n",
    "       inversion Heq1. subst t1. subst t2. subst t3.\n",
    "       inversion Heq2. reflexivity.\n",
    "Qed.\n",
    "\n",
    "Lemma succ_not_circular_sol : forall t,\n",
    "  t <> tm_succ t.\n",
    "Proof.\n",
    "  intros t. induction t.\n",
    "    intros Heq. inversion Heq.\n",
    "    intros Heq. inversion Heq.\n",
    "    intros Heq. inversion Heq.\n",
    "    intros Heq. inversion Heq.\n",
    "    intros Heq. inversion Heq. destruct IHt. apply H0.\n",
    "    intros Heq. inversion Heq.\n",
    "    intros Heq. inversion Heq.\n",
    "Qed.\n",
    "\n",
    "Lemma inversion_exercise_sol : forall t,\n",
    "  normal_form t ->\n",
    "  eval_many (tm_pred t) tm_zero ->\n",
    "  nvalue t.\n",
    "Proof.\n",
    "  intros t Hnf Hm. inversion Hm. subst. inversion H.\n",
    "    apply n_zero.\n",
    "    apply n_succ. apply H2.\n",
    "    destruct Hnf. exists t'0. apply H2.\n",
    "Qed.\n",
    "\n",
    "Lemma contradictory_equalities_exercise_sol :\n",
    "  (exists t, exists u, exists v,\n",
    "    value t /\\\n",
    "    t = tm_succ u /\\\n",
    "    u = tm_pred v) ->\n",
    "  eval tm_true tm_false.\n",
    "Proof.\n",
    "  intros [ t [ u [ v [ [ Hb | Hn ] [ eq1 eq2 ] ] ] ] ].\n",
    "    destruct Hb.\n",
    "      inversion eq1.\n",
    "      inversion eq1.\n",
    "    destruct Hn.\n",
    "      inversion eq1.\n",
    "      inversion eq1. subst t. subst u. inversion Hn.\n",
    "Qed.\n",
    "\n",
    "Lemma eval_fact_exercise_sol : forall t1 t2,\n",
    "  eval (tm_iszero (tm_pred t1)) t2 ->\n",
    "  eval t2 tm_false ->\n",
    "  exists u, t1 = tm_succ u.\n",
    "Proof.\n",
    "  intros t1 t2 He1 He2. inversion He1. subst t2.\n",
    "   inversion He2. subst t'.\n",
    "   inversion H0. exists (tm_succ t0). reflexivity.\n",
    "Qed.\n",
    "\n",
    "Lemma normal_form_if_sol : forall t1 t2 t3,\n",
    "  normal_form (tm_if t1 t2 t3) ->\n",
    "  t1 <> tm_true /\\ t1 <> tm_false /\\ normal_form t1.\n",
    "Proof.\n",
    "  intros t1 t2 t3 Hnf. destruct t1.\n",
    "    destruct Hnf. exists t2. apply e_iftrue.\n",
    "    destruct Hnf. exists t3. apply e_iffalse.\n",
    "    split.\n",
    "      intros Heq. inversion Heq.\n",
    "      split.\n",
    "        intros Heq. inversion Heq.\n",
    "        intros [t' He]. destruct Hnf. exists (tm_if t' t2 t3).\n",
    "         apply e_if. apply He.\n",
    "    split.\n",
    "      intros Heq. inversion Heq.\n",
    "      split.\n",
    "        intros Heq. inversion Heq.\n",
    "        intros [t' He]. inversion He.\n",
    "    split.\n",
    "      intros Heq. inversion Heq.\n",
    "      split.\n",
    "        intros Heq. inversion Heq.\n",
    "        intros [t' He]. destruct Hnf. exists (tm_if t' t2 t3).\n",
    "         apply e_if. apply He.\n",
    "    split.\n",
    "      intros Heq. inversion Heq.\n",
    "      split.\n",
    "        intros Heq. inversion Heq.\n",
    "        intros [t' He]. destruct Hnf. exists (tm_if t' t2 t3).\n",
    "         apply e_if. apply He.\n",
    "    split.\n",
    "      intros Heq. inversion Heq.\n",
    "      split.\n",
    "        intros Heq. inversion Heq.\n",
    "        intros [t' He]. destruct Hnf. exists (tm_if t' t2 t3).\n",
    "         apply e_if. apply He.\n",
    "Qed.\n",
    "\n",
    "Lemma full_eval_from_value_sol : forall v w,\n",
    "  value v ->\n",
    "  full_eval v w ->\n",
    "  v = w.\n",
    "Proof.\n",
    "  intros v w Hv Hf. induction Hf.\n",
    "    reflexivity.\n",
    "    destruct Hv as [ Hb | Hn ].\n",
    "      inversion Hb.\n",
    "      inversion Hn.\n",
    "    destruct Hv as [ Hb | Hn ].\n",
    "      inversion Hb.\n",
    "      inversion Hn.\n",
    "    rewrite -> IHHf.\n",
    "      reflexivity.\n",
    "      right. apply value_succ_nvalue. apply Hv.\n",
    "    destruct Hv as [ Hb | Hn ].\n",
    "      inversion Hb.\n",
    "      inversion Hn.\n",
    "    destruct Hv as [ Hb | Hn ].\n",
    "      inversion Hb.\n",
    "      inversion Hn.\n",
    "    destruct Hv as [ Hb | Hn ].\n",
    "      inversion Hb.\n",
    "      inversion Hn.\n",
    "    destruct Hv as [ Hb | Hn ].\n",
    "      inversion Hb.\n",
    "      inversion Hn.\n",
    "Qed.\n",
    "\n",
    "Lemma value_is_normal_form_sol : forall v,\n",
    "  value v ->\n",
    "  normal_form v.\n",
    "Proof.\n",
    "  intros v [ Hb | Hn ] [ t He ].\n",
    "    destruct Hb; inversion He.\n",
    "    generalize dependent t.\n",
    "     induction Hn; intros u He; inversion He; subst.\n",
    "      destruct (IHHn t'). apply H0.\n",
    "Qed.\n",
    "\n",
    "Theorem eval_deterministic_sol : forall t t' t'',\n",
    "  eval t t' ->\n",
    "  eval t t'' ->\n",
    "  t' = t''.\n",
    "Proof.\n",
    "  intros t t' t'' He1. generalize dependent t''.\n",
    "   induction He1; intros t'' He2; inversion He2; subst.\n",
    "    reflexivity.\n",
    "    inversion H3.\n",
    "    reflexivity.\n",
    "    inversion H3.\n",
    "    inversion He1.\n",
    "    inversion He1.\n",
    "    rewrite -> (IHHe1 t1'0).\n",
    "      reflexivity.\n",
    "      apply H3.\n",
    "    rewrite -> (IHHe1 t'0).\n",
    "      reflexivity.\n",
    "      apply H0.\n",
    "    reflexivity.\n",
    "    inversion H0.\n",
    "    reflexivity.\n",
    "    assert (normal_form (tm_succ t)) as Hnf.\n",
    "      apply nvalue_is_normal_form. apply n_succ. apply H.\n",
    "     destruct Hnf. exists t'. apply H1.\n",
    "    inversion He1.\n",
    "    assert (normal_form (tm_succ t'')) as Hnf.\n",
    "      apply nvalue_is_normal_form. apply n_succ. apply H0.\n",
    "     destruct Hnf. exists t'. apply He1.\n",
    "    rewrite -> (IHHe1 t'0).\n",
    "      reflexivity.\n",
    "      apply H0.\n",
    "    reflexivity.\n",
    "    inversion H0.\n",
    "    reflexivity.\n",
    "    assert (normal_form (tm_succ t)) as Hnf.\n",
    "      apply nvalue_is_normal_form. apply n_succ. apply H.\n",
    "     destruct Hnf. exists t'. apply H1.\n",
    "    inversion He1.\n",
    "    assert (normal_form (tm_succ t0)) as Hnf.\n",
    "      apply nvalue_is_normal_form. apply n_succ. apply H0.\n",
    "     destruct Hnf. exists t'. apply He1.\n",
    "    rewrite -> (IHHe1 t'0).\n",
    "      reflexivity.\n",
    "      apply H0.\n",
    "Qed.\n",
    "\n",
    "Lemma eval_full_eval_sol : forall t t' v,\n",
    "  eval t t' ->\n",
    "  full_eval t' v ->\n",
    "  full_eval t v.\n",
    "Proof.\n",
    "  intros t t' v He. generalize dependent v. induction He.\n",
    "    intros v Hf. apply f_iftrue.\n",
    "      apply f_value. left. apply b_true.\n",
    "      apply Hf.\n",
    "    intros v Hf. apply f_iffalse.\n",
    "      apply f_value. left. apply b_false.\n",
    "      apply Hf.\n",
    "    intros v Hf. inversion Hf.\n",
    "      subst. inversion H.\n",
    "        inversion H0.\n",
    "        inversion H0.\n",
    "      subst. apply f_iftrue.\n",
    "        apply IHHe. apply H3.\n",
    "        apply H4.\n",
    "      subst. apply f_iffalse.\n",
    "        apply IHHe. apply H3.\n",
    "        apply H4.\n",
    "    intros v Hf. inversion Hf.\n",
    "      subst. apply f_succ.\n",
    "        apply value_succ_nvalue. apply H.\n",
    "        apply IHHe. apply f_value. right.\n",
    "             apply value_succ_nvalue. apply H.\n",
    "      subst. apply f_succ.\n",
    "        apply H0.\n",
    "        apply IHHe. apply H1.\n",
    "    intros v Hf. inversion Hf. apply f_predzero.\n",
    "         apply f_value. right. apply n_zero.\n",
    "    intros v Hf. assert (t = v).\n",
    "      apply full_eval_from_value_sol.\n",
    "        right. apply H.\n",
    "        apply Hf.\n",
    "      subst v. apply f_predsucc.\n",
    "        apply H.\n",
    "        apply f_succ.\n",
    "          apply H.\n",
    "          apply Hf.\n",
    "    intros v Hf. inversion Hf.\n",
    "      subst. destruct H as [ Hb | Hn ].\n",
    "        inversion Hb.\n",
    "        inversion Hn.\n",
    "      subst. apply f_predzero. apply IHHe. apply H0.\n",
    "      subst. apply f_predsucc.\n",
    "        apply H0.\n",
    "        apply IHHe. apply H1.\n",
    "    intros v Hf. inversion Hf. apply f_iszerozero.\n",
    "         apply f_value. right. apply n_zero.\n",
    "    intros v Hf. inversion Hf.\n",
    "         apply f_iszerosucc with (v := t).\n",
    "      apply H.\n",
    "      apply f_value. right. apply n_succ. apply H.\n",
    "    intros v Hf. inversion Hf.\n",
    "      subst. destruct H as [ Hb | Hn ].\n",
    "        inversion Hb.\n",
    "        inversion Hn.\n",
    "      subst. apply f_iszerozero. apply IHHe. apply H0.\n",
    "      subst. apply f_iszerosucc with (v := v0).\n",
    "        apply H0.\n",
    "        apply IHHe. apply H1.\n",
    "Qed.\n",
    "\n",
    "Lemma full_eval_complete_sol : forall t v,\n",
    "  value v ->\n",
    "  eval_many t v ->\n",
    "  full_eval t v.\n",
    "Proof.\n",
    "  intros t v Hv Hm. induction Hm.\n",
    "    apply f_value. apply Hv.\n",
    "    apply eval_full_eval_sol with (t' := t').\n",
    "      apply H.\n",
    "      apply IHHm. apply Hv.\n",
    "Qed.\n",
    "\n",
    "Lemma pred_not_circular_sol : forall t,\n",
    "  t <> tm_pred t.\n",
    "Proof.\n",
    "  intros t H. induction t; inversion H; auto.\n",
    "Qed.\n",
    "\n",
    "Lemma m_succ_sol : forall t u,\n",
    "  eval_many t u ->\n",
    "  eval_many (tm_succ t) (tm_succ u).\n",
    "Proof.\n",
    "  intros t u Hm.\n",
    "   induction Hm; eauto using m_refl, m_step, e_succ.\n",
    "Qed.\n",
    "\n",
    "Lemma m_pred_sol : forall t u,\n",
    "  eval_many t u ->\n",
    "  eval_many (tm_pred t) (tm_pred u).\n",
    "Proof.\n",
    "  intros t u Hm.\n",
    "   induction Hm; eauto using m_refl, m_step, e_pred.\n",
    "Qed.\n",
    "\n",
    "Fixpoint interp (t : tm) {struct t} : tm :=\n",
    "  match t with\n",
    "  | tm_true => tm_true\n",
    "  | tm_false => tm_false\n",
    "  | tm_if t1 t2 t3 =>\n",
    "      match interp t1 with\n",
    "      | tm_true => interp t2\n",
    "      | tm_false => interp t3\n",
    "      | t4 => tm_if t4 t2 t3\n",
    "      end\n",
    "  | tm_zero => tm_zero\n",
    "  | tm_succ t1 => tm_succ (interp t1)\n",
    "  | tm_pred t1 =>\n",
    "      match interp t1 with\n",
    "      | tm_zero => tm_zero\n",
    "      | tm_succ t2 =>\n",
    "          match tm_to_nat t2 with\n",
    "          | some_nat _ => t2\n",
    "          | no_nat => tm_pred (tm_succ t2)\n",
    "          end\n",
    "      | t2 => tm_pred t2\n",
    "      end\n",
    "  | tm_iszero t1 =>\n",
    "      match interp t1 with\n",
    "      | tm_zero => tm_true\n",
    "      | tm_succ t2 =>\n",
    "          match tm_to_nat t2 with\n",
    "          | some_nat _ => tm_false\n",
    "          | no_nat => tm_iszero (tm_succ t2)\n",
    "          end\n",
    "      | t2 => tm_iszero t2\n",
    "      end\n",
    "  end.\n",
    "\n",
    "Lemma tm_to_bool_dom_includes_bvalue_sol : forall bv,\n",
    "  bvalue bv -> exists b, tm_to_bool bv = some_bool b.\n",
    "Proof.\n",
    "  intros bv H. destruct H.\n",
    "    exists true. reflexivity.\n",
    "    exists false. reflexivity.\n",
    "Qed.\n",
    "\n",
    "Lemma tm_to_bool_dom_only_bvalue_sol : forall bv b,\n",
    "  tm_to_bool bv = some_bool b -> bvalue bv.\n",
    "Proof.\n",
    "  intros bv b Heq. destruct bv.\n",
    "    apply b_true.\n",
    "    apply b_false.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "Qed.\n",
    "\n",
    "Lemma tm_to_nat_dom_includes_nvalue_sol : forall v,\n",
    "  nvalue v -> exists n, tm_to_nat v = some_nat n.\n",
    "Proof.\n",
    "  intros v Hnv. induction Hnv.\n",
    "    exists O. reflexivity.\n",
    "    destruct IHHnv as [ n Heq ]. exists (S n).\n",
    "     simpl. rewrite -> Heq. reflexivity.\n",
    "Qed.\n",
    "\n",
    "Lemma tm_to_nat_dom_only_nvalue_sol : forall v n,\n",
    "  tm_to_nat v = some_nat n -> nvalue v.\n",
    "Proof.\n",
    "  intros v. induction v; intros n Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "    apply n_zero.\n",
    "    apply n_succ.\n",
    "     simpl in Heq. destruct (tm_to_nat v).\n",
    "      inversion Heq. eapply IHv. reflexivity.\n",
    "      inversion Heq.\n",
    "    inversion Heq.\n",
    "    inversion Heq.\n",
    "Qed.\n",
    "\n",
    "Lemma interp_reduces_sol : forall t,\n",
    "  eval_many t (interp t).\n",
    "Proof.\n",
    "  intros t. induction t.\n",
    "    apply m_refl.\n",
    "    apply m_refl.\n",
    "    simpl. destruct (interp t1).\n",
    "      eapply m_trans.\n",
    "        apply m_if. apply IHt1.\n",
    "        eapply m_trans.\n",
    "          eapply m_one. apply e_iftrue.\n",
    "          apply IHt2.\n",
    "      eapply m_trans.\n",
    "        apply m_if. apply IHt1.\n",
    "        eapply m_trans.\n",
    "          eapply m_one. apply e_iffalse.\n",
    "          apply IHt3.\n",
    "      apply m_if. apply IHt1.\n",
    "      apply m_if. apply IHt1.\n",
    "      apply m_if. apply IHt1.\n",
    "      apply m_if. apply IHt1.\n",
    "      apply m_if. apply IHt1.\n",
    "    apply m_refl.\n",
    "    simpl. apply m_succ. apply IHt.\n",
    "    simpl. destruct (interp t).\n",
    "      apply m_pred. apply IHt.\n",
    "      apply m_pred. apply IHt.\n",
    "      apply m_pred. apply IHt.\n",
    "      eapply m_trans.\n",
    "        apply m_pred. apply IHt.\n",
    "        apply m_one. apply e_predzero.\n",
    "      remember (tm_to_nat t0) as x. destruct x.\n",
    "        eapply m_trans.\n",
    "          apply m_pred. apply IHt.\n",
    "          apply m_one. apply e_predsucc.\n",
    "           eapply tm_to_nat_dom_only_nvalue.\n",
    "           rewrite <- Heqx. reflexivity.\n",
    "        apply m_pred. apply IHt.\n",
    "      apply m_pred. apply IHt.\n",
    "      apply m_pred. apply IHt.\n",
    "    simpl. destruct (interp t).\n",
    "      apply m_iszero. apply IHt.\n",
    "      apply m_iszero. apply IHt.\n",
    "      apply m_iszero. apply IHt.\n",
    "      eapply m_trans.\n",
    "        apply m_iszero. apply IHt.\n",
    "        apply m_one. apply e_iszerozero.\n",
    "      remember (tm_to_nat t0) as x. destruct x.\n",
    "        eapply m_trans.\n",
    "          apply m_iszero. apply IHt.\n",
    "          apply m_one. apply e_iszerosucc.\n",
    "            eapply tm_to_nat_dom_only_nvalue.\n",
    "            rewrite <- Heqx. reflexivity.\n",
    "        apply m_iszero. apply IHt.\n",
    "      apply m_iszero. apply IHt.\n",
    "      apply m_iszero. apply IHt.\n",
    "Qed.\n",
    "\n",
    "Lemma interp_fully_reduces_sol : forall t,\n",
    "  normal_form (interp t).\n",
    "Proof.\n",
    "  induction t; intros [t' H].\n",
    "    inversion H.\n",
    "    inversion H.\n",
    "    simpl in H. destruct (interp t1).\n",
    "      destruct IHt2. eauto.\n",
    "      destruct IHt3. eauto.\n",
    "      destruct IHt1. inversion H. eauto.\n",
    "      destruct IHt1. inversion H. eauto.\n",
    "      destruct IHt1. inversion H. eauto.\n",
    "      destruct IHt1. inversion H. eauto.\n",
    "      destruct IHt1. inversion H. eauto.\n",
    "    inversion H.\n",
    "    destruct IHt. inversion H. eauto.\n",
    "    simpl in H. destruct (interp t).\n",
    "      inversion H. inversion H1.\n",
    "      inversion H. inversion H1.\n",
    "      inversion H. destruct IHt. eauto.\n",
    "      inversion H.\n",
    "      remember (tm_to_nat t0) as x. destruct x.\n",
    "        destruct IHt. exists (tm_succ t').\n",
    "         apply e_succ. apply H.\n",
    "        inversion H; subst.\n",
    "          destruct (tm_to_nat_dom_includes_nvalue t')\n",
    "           as [n Heq].\n",
    "            apply H1.\n",
    "            rewrite <- Heqx in Heq. inversion Heq.\n",
    "          destruct IHt. eauto.\n",
    "      inversion H. destruct IHt. eauto.\n",
    "      inversion H. destruct IHt. eauto.\n",
    "    simpl in H. destruct (interp t).\n",
    "      inversion H. inversion H1.\n",
    "      inversion H. inversion H1.\n",
    "      inversion H. destruct IHt. eauto.\n",
    "      inversion H.\n",
    "      remember (tm_to_nat t0) as x. destruct x.\n",
    "        inversion H.\n",
    "        inversion H; subst.\n",
    "          destruct (tm_to_nat_dom_includes_nvalue t0)\n",
    "           as [n Heq].\n",
    "            apply H1.\n",
    "            rewrite <- Heqx in Heq. inversion Heq.\n",
    "          inversion H. destruct IHt. eauto.\n",
    "      inversion H. destruct IHt. eauto.\n",
    "      inversion H. destruct IHt. eauto.\n",
    "Qed."
   ]
  }
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