(** * Smallstep: Small-step operational semantics *) (* Version of 9/10/2009 *) Require Export Hoare. Require Import Relations. (** The evaluators we have seen so far (e.g., the ones for [aexp]s, [bexp]s, and commands) have been formulated in a "big-step" style -- they specify how a given expression can be evaluated to its final value (or a command plus a store to a final store) "all in one big step." This style is simple and natural for many purposes, but it has some shortcomings. In particular, suppose we enriched the language of while programs with boolean variables in addition to the numeric ones. In this language, a command might _fail_ to map a given starting state to any ending state for two quite different reasons: either because the execution gets into an infinite loop or because, at some point, the program tries to do an operation that makes no sense, such as taking the successor of a boolean variable, and none of the evaluation rules can be applied. These two outcomes -- nontermination vs. getting stuck in an erroneous configuration -- need to be treated differently: we want to allow the first (permitting the possibility of infinite loops is the price we pay for the convenience of being able to program with general looping constructs like [while]) but prevent the second, for example by adding some form of _typechecking_ to the language. (Indeed, this will be a major topic for the rest of the course.) As a first step, it is useful to introduce a different way of defining how programs behave -- replacing the "big-step" [eval] relation with a "small-step" relation that specifies, for a given program, how just the FIRST atomic step of computation is to be performed. *) (* ########################################################### *) (** * A toy language *) (** To save space in the discussion, let's work with an incredibly simple language containing just constants and addition. *) Inductive tm : Type := | tm_const : nat -> tm | tm_plus : tm -> tm -> tm. Tactic Notation "tm_cases" tactic(first) tactic(c) := first; [ c "tm_const" | c "tm_plus" ]. Module SimpleArith1. (** Here is a standard big-step evaluator for this language, in exactly the same style as we've been using up to this point. *) Inductive eval : tm -> nat -> Prop := | E_Const : forall n, eval (tm_const n) n | E_Plus : forall t1 t2 n1 n2, eval t1 n1 -> eval t2 n2 -> eval (tm_plus t1 t2) (plus n1 n2). End SimpleArith1. (** Here is a slight variation (still in "big-step" style) where the final result of evaluating a term is also a term. *) Inductive eval : tm -> tm -> Prop := | E_Const : forall n1, eval (tm_const n1) (tm_const n1) | E_Plus : forall t1 n1 t2 n2, eval t1 (tm_const n1) -> eval t2 (tm_const n2) -> eval (tm_plus t1 t2) (tm_const (plus n1 n2)). Tactic Notation "eval_cases" tactic(first) tactic(c) := first; [ c "E_Const" | c "E_Plus" ]. Module SimpleArith2. (** Now, here is the small-step variant. *) Inductive step : tm -> tm -> Prop := | ST_PlusConstConst : forall n1 n2, step (tm_plus (tm_const n1) (tm_const n2)) (tm_const (plus n1 n2)) | ST_Plus1 : forall t1 t1' t2, (step t1 t1') -> step (tm_plus t1 t2) (tm_plus t1' t2) | ST_Plus2 : forall n1 t2 t2', (step t2 t2') -> step (tm_plus (tm_const n1) t2) (tm_plus (tm_const n1) t2'). (** A few things to notice: - We are defining just a single reduction step, in which one [tm_plus] node is replaced by its value. - Each step finds the _leftmost_ [tm_plus] node that is "ready to go" (both of its operands are constants) and reduces it. The first rule tells how to reduce this [tm_plus] node itself; the other two rules tell how to find it. - A term that is just a constant cannot take a step. *) Tactic Notation "step_cases" tactic(first) tactic(c) := first; [ c "ST_PlusConstConst" | c "ST_Plus1" | c "ST_Plus2" ]. (** A couple of examples of reasoning with the [step] relation... *) (** If [t1] can take a step to [t1'], then [tm_plus t1 t2] steps to [plus t1' t2]: *) Example test_step_1 : step (tm_plus (tm_plus (tm_const 0) (tm_const 3)) (tm_plus (tm_const 2) (tm_const 4))) (tm_plus (tm_const (plus 0 3)) (tm_plus (tm_const 2) (tm_const 4))). Proof. apply ST_Plus1. apply ST_PlusConstConst. Qed. (** **** Exercise: 2 stars (test_step_2) *) (** Right-hand sides of sums can take a step only when the left-hand side is finished: if [t2] can take a step to [t2'], then [tm_plus (tm_const n) t2] steps to [tm_plus (tm_const n) t2']: *) Example test_step_2 : step (tm_plus (tm_const 0) (tm_plus (tm_const 2) (tm_plus (tm_const 0) (tm_const 3)))) (tm_plus (tm_const 0) (tm_plus (tm_const 2) (tm_const (plus 0 3)))). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** One interesting property of the [step] relation is that, like the evaluation relation for our language of While programs, it is _deterministic_: for each [t], there is at most one [t'] such that [step t t'] is provable. Formally, this is the same as saying that [step] is a partial function. *) (** Proof sketch: We must show that if [x] steps to both [y1] and [y2] then [y1] and [y2] are equal. Consider the last rules used in the derivations of [step x y1] and [step x y2]. - If both are [ST_PlusConstConst], the result is immediate. - It cannot happen that one is [ST_PlusConstConst] and the other is [ST_Plus1] or [ST_Plus2], since this would imply that [x] has the form [tm_plus t1 t2] where both [t1] and [t2] are constants (by [ST_PlusConstConst]) AND one of [t1] or [t2] has the form [tm_plus ...]. - Similarly, it cannot happen that one is [ST_Plus1] and the other is [ST_Plus2], since this would imply that [x] has the form [tm_plus t1 t2] where [t1] has both the form [tm_plus t1 t2] and the form [tm_const n]. - The cases when both derivations end with [ST_Plus1] or [ST_Plus2] follow by the induction hypothesis. *) Theorem step_deterministic : partial_function step. Proof. unfold partial_function. intros x y1 y2 Hy1 Hy2. generalize dependent y2. step_cases (induction Hy1) Case. Case "ST_PlusConstConst". intros y2 Hy2. step_cases (inversion Hy2) SCase. SCase "ST_PlusConstConst". reflexivity. SCase "ST_Plus1". inversion H2. SCase "ST_Plus2". inversion H2. Case "ST_Plus1". intros y2 Hy2. step_cases (inversion Hy2) SCase. SCase "ST_PlusConstConst". rewrite <- H0 in Hy1. inversion Hy1. SCase "ST_Plus1". rewrite <- (IHHy1 t1'0). reflexivity. assumption. SCase "ST_Plus2". rewrite <- H in Hy1. inversion Hy1. Case "ST_Plus2". intros y2 Hy2. step_cases (inversion Hy2) SCase. SCase "ST_PlusConstConst". rewrite <- H1 in Hy1. inversion Hy1. SCase "ST_Plus1". inversion H2. SCase "ST_Plus2". rewrite <- (IHHy1 t2'0). reflexivity. assumption. Qed. End SimpleArith2. (* ########################################################### *) (** ** Values *) (** Before we move on, let's take a moment to slightly generalize the way we state the definition of single-step reduction. It is useful to think of the [step] relation as defining a sort of _abstract machine_ for evaluating programs: - At any moment, the _state_ of the machine is a term. - A _step_ of the machine is an atomic unit of computation -- a single "add" operation, in the case of the present tiny programming language. - The _final states_ of the machine are ones where there is no more computation to be done. We can then think about "executing" a term [t] as follows: - Take [t] as the starting state of the machine. - Repeatedly use the [step] relation to find a sequence of machine states such that each steps to the next. - When no more reduction is possible, "read out" the final state of the machine as the result of execution. Intuitively, it is clear that the final states of the machine are always terms of the form [tm_const n] for some [n]. We call such terms _values_. *) Inductive value : tm -> Prop := v_const : forall n, value (tm_const n). (** Having introduced the idea of values, we can use it in the definition of the [step] relation to write [ST_Plus2] rule in a slightly more intuitive way: *) Inductive step : tm -> tm -> Prop := | ST_PlusConstConst : forall n1 n2, step (tm_plus (tm_const n1) (tm_const n2)) (tm_const (plus n1 n2)) | ST_Plus1 : forall t1 t1' t2, (step t1 t1') -> step (tm_plus t1 t2) (tm_plus t1' t2) | ST_Plus2 : forall v1 t2 t2', (value v1) (* <----- *) -> (step t2 t2') -> step (tm_plus v1 t2) (tm_plus v1 t2'). Tactic Notation "step_cases" tactic(first) tactic(c) := first; [ c "ST_PlusConstConst" | c "ST_Plus1" | c "ST_Plus2" ]. (** **** Exercise: 3 stars (redo_determinacy) *) (** As a sanity check on this change, let's re-verify determinacy Proof sketch: We must show that if [x] steps to both [y1] and [y2] then [y1] and [y2] are equal. Consider the final rules used in the derivations of [step x y1] and [step x y2]. - If both are [ST_PlusConstConst], the result is immediate. - It cannot happen that one is [ST_PlusConstConst] and the other is [ST_Plus1] or [ST_Plus2], since this would imply that [x] has the form [tm_plus t1 t2] where both [t1] and [t2] are constants (by [ST_PlusConstConst]) AND one of [t1] or [t2] has the form [tm_plus ...]. - Similarly, it cannot happen that one is [ST_Plus1] and the other is [ST_Plus2], since this would imply that [x] has the form [tm_plus t1 t2] where [t1] both has the form [tm_plus t1 t2] and is a value (hence has the form [tm_const n]). - The cases when both derivations end with [ST_Plus1] or [ST_Plus2] follow by the induction hypothesis. [] *) (** Most of this proof is the same as the one above. But to get maximum benefit from the exercise you should try to write it from scratch and just use the earlier one if you get stuck. *) Theorem step_deterministic : partial_function step. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ########################################################### *) (** ** Normal forms *) (** A fundamental property of this language is that every term is either a value or it can "make progress" by stepping to some other term. This property is called _progress_. Proof sketch: By induction on [t]. - If [t] is a constant, then it is a [value]. - If [t = tm_plus t1 t2], then by the IH [t1] and [t2] are either values or can take steps under [step]. - If [t1] and [t2] are both values, then [t] can take a step, by [ST_PlusConstConst]. - If [t1] is a value and [t2] can take a step, then so can [t], by [ST_Plus2]. - If [t1] can take a step, then so can [t], by [ST_Plus1]. [] *) Theorem progress : forall t, value t \/ (exists t', step t t'). Proof. tm_cases (induction t) Case. Case "tm_const". left. apply v_const. Case "tm_plus". right. inversion IHt1. SCase "l". inversion IHt2. SSCase "l". inversion H. inversion H0. exists (tm_const (plus n n0)). apply ST_PlusConstConst. SSCase "r". inversion H0 as [t' H1]. exists (tm_plus t1 t'). apply ST_Plus2. apply H. apply H1. SCase "r". inversion H as [t' H0]. exists (tm_plus t' t2). apply ST_Plus1. apply H0. Qed. (** This property can be extended to tell us something very interesting about [value]s: they are exactly the terms that _cannot_ make progress in this sense. To state this fact, let's begin by giving a name to terms that cannot make progress: We'll call them _normal forms_. *) Definition normal_form (X:Type) (R:relation X) (t:X) : Prop := ~ exists t', R t t'. Implicit Arguments normal_form [X]. (** We've actually defined what it is to be a normal form for an arbitrary relation [R] over an arbitrary set [X], not just for the particular reduction relation over terms that we are interested in at the moment. We'll re-use the same terminology for talking about other relations later in the course. *) (** We can use this terminology to generalize the observation we made in the progress theorem: normal forms and values are actually the same thing. Note that we state and prove this result as two different lemmas, rather than using an if-and-only-if ([<->]). That's because it will be easier to apply the separate lemmas later on; as noted before, Coq's facilities for dealing "in-line" with [<->] statements are a little awkward. *) Lemma value_is_nf : forall t, value t -> normal_form step t. Proof. intros t H. unfold normal_form. intros contra. inversion H. rewrite <- H0 in contra. destruct contra as [t' P]. inversion P. Qed. (** Proof sketch: This is a corollary of [progress]. *) Lemma nf_is_value : forall t, normal_form step t -> value t. Proof. intros t H. unfold normal_form in H. assert (value t \/ exists t', step t t') as G. SCase "Proof of assertion". apply progress. inversion G. SCase "l". apply H0. SCase "r". apply ex_falso_quodlibet. apply H. assumption. Qed. (** Why are these last two facts interesting? For two reasons: - Because [value] is a syntactic concept -- it is a defined by looking at the form of a term -- while [normal_form] is a semantic one -- it is defined by looking at how the term steps. Is it not obvious that these concepts should coincide. - Indeed, there are lots of languages in which the concepts of normal form and value do _not_ coincide. Let's examine how this can happen... *) (* -------------------------------------------------- *) (** We might, for example, accidentally define [value] so that it includes some terms that are not finished reducing. *) Module Temp1. (* Open an inner module so we can redefine value and step. *) Inductive value : tm -> Prop := | v_const : forall n, value (tm_const n) | v_funny : forall t1 n2, (* <---- *) value (tm_plus t1 (tm_const n2)). Inductive step : tm -> tm -> Prop := | ST_PlusConstConst : forall n1 n2, step (tm_plus (tm_const n1) (tm_const n2)) (tm_const (plus n1 n2)) | ST_Plus1 : forall t1 t1' t2, (step t1 t1') -> step (tm_plus t1 t2) (tm_plus t1' t2) | ST_Plus2 : forall v1 t2 t2', (value v1) -> (step t2 t2') -> step (tm_plus v1 t2) (tm_plus v1 t2'). (** **** Exercise: 3 stars (value_not_same_as_normal_form) *) Lemma value_not_same_as_normal_form : exists t, value t /\ ~ normal_form step t. Proof. (* FILL IN HERE *) Admitted. (** [] *) End Temp1. (* -------------------------------------------------- *) (** Alternatively, we might accidentally define [step] so that it permits something designated as a value to reduce further. *) Module Temp2. Inductive value : tm -> Prop := | v_const : forall n, value (tm_const n). Inductive step : tm -> tm -> Prop := | ST_Funny : forall n, (* <---- *) step (tm_const n) (tm_plus (tm_const n) (tm_const 0)) | ST_PlusConstConst : forall n1 n2, step (tm_plus (tm_const n1) (tm_const n2)) (tm_const (plus n1 n2)) | ST_Plus1 : forall t1 t1' t2, (step t1 t1') -> step (tm_plus t1 t2) (tm_plus t1' t2) | ST_Plus2 : forall v1 t2 t2', (value v1) -> (step t2 t2') -> step (tm_plus v1 t2) (tm_plus v1 t2'). (** **** Exercise: 3 stars (value_not_same_as_normal_form) *) Lemma value_not_same_as_normal_form : exists t, value t /\ ~ normal_form step t. Proof. (* FILL IN HERE *) Admitted. (** [] *) End Temp2. (* -------------------------------------------------- *) (** Finally, we might accidentally define [value] and [step] so that there is some term that is not a value but that cannot take a step in the [step] relation. Such terms are said to be _stuck_. *) Module Temp3. Inductive value : tm -> Prop := | v_const : forall n, value (tm_const n). Inductive step : tm -> tm -> Prop := | ST_PlusConstConst : forall n1 n2, step (tm_plus (tm_const n1) (tm_const n2)) (tm_const (plus n1 n2)) | ST_Plus1 : forall t1 t1' t2, (step t1 t1') -> step (tm_plus t1 t2) (tm_plus t1' t2). (** Note that ST_Plus2 is missing. *) (** **** Exercise: 3 stars (value_not_same_as_normal_form') *) Lemma value_not_same_as_normal_form : exists t, ~ value t /\ normal_form step t. Proof. (* FILL IN HERE *) Admitted. (** [] *) End Temp3. (* ########################################################### *) (** ** Exercises *) Module Temp4. (** Here is another very simple language whose terms, instead of being just plus and numbers, are just the booleans true and false and a conditional expression... *) Inductive tm : Type := | tm_true : tm | tm_false : tm | tm_if : tm -> tm -> tm -> tm. Inductive value : tm -> Prop := | v_true : value tm_true | v_false : value tm_false. Inductive step : tm -> tm -> Prop := | ST_IfTrue : forall t1 t2, step (tm_if tm_true t1 t2) t1 | ST_IfFalse : forall t1 t2, step (tm_if tm_false t1 t2) t2 | ST_If : forall t1 t1' t2 t3, step t1 t1' -> step (tm_if t1 t2 t3) (tm_if t1' t2 t3). (** **** Exercise: 1 star (smallstep_bools) *) (** Which of the following propositions are provable? (This is just a thought exercise, but for an extra challenge feel free to prove your answers in Coq.) *) Definition bool_step_prop1 := step tm_false tm_false. (* FILL IN HERE *) Definition bool_step_prop2 := step (tm_if tm_true (tm_if tm_true tm_true tm_true) (tm_if tm_false tm_false tm_false)) tm_true. (* FILL IN HERE *) Definition bool_step_prop3 := step (tm_if (tm_if tm_true tm_true tm_true) (tm_if tm_true tm_true tm_true) tm_false) (tm_if tm_true (tm_if tm_true tm_true tm_true) tm_false). (* FILL IN HERE *) (** [] *) (** **** Exercise: 3 stars (progress_bool) *) (** Just as we proved a progress theorem for plus expressions, we can do so for boolean expressions, as well. *) Theorem progress : forall t, value t \/ (exists t', step t t'). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 2 stars, optional (step_deterministic) *) Theorem step_deterministic : partial_function step. Proof. (* FILL IN HERE *) Admitted. (** [] *) Module Temp5. (** **** Exercise: 2 stars (smallstep_bool_shortcut) *) (** Suppose we want to add a "short circuit" to the step relation for boolean expressions, so that it can recognize when the [then] and [else] branches of a conditional are the same value (either [tm_true] or [tm_false]) and reduce the whole conditional to this value in a single step, even if the guard has not yet been reduced to a value. For example, we would like this proposition to be provable: *) (** Write an extra clause for the step relation that achieves this effect and prove [bool_step_prop4]. *) Inductive step : tm -> tm -> Prop := | ST_IfTrue : forall t1 t2, step (tm_if tm_true t1 t2) t1 | ST_IfFalse : forall t1 t2, step (tm_if tm_false t1 t2) t2 | ST_If : forall t1 t1' t2 t3, step t1 t1' -> step (tm_if t1 t2 t3) (tm_if t1' t2 t3) (* FILL IN HERE *) . (** [] *) (** **** Exercise: 2 stars (bool_step_prop4_holds) *) (** To check that your previous answer is correct, prove that the following step is now possible. *) Definition bool_step_prop4 := step (tm_if (tm_if tm_true tm_true tm_true) tm_false tm_false) tm_false. Example bool_step_prop4_holds : bool_step_prop4. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars (properties_of_altered_step) *) (** It can be shown that the determinism and progress theorems for the step relation in the lecture notes also hold for the definition of step given above. After we add the clause [ST_ShortCircuit]... - Does step_deterministic still hold? Write yes or no and briefly (1 sentence) explain your answer. Optional exercise: prove your answer correct in Coq. *) (* FILL IN HERE *) (** - Does a progress theorem hold? Write yes or no and briefly (1 sentence) explain your answer. Optional exercise: prove your answer correct in Coq. *) (* FILL IN HERE *) (** - In general, is there any way we could cause progress to fail if we took away one or more constructors from the original step relation? Write yes or no and briefly (1 sentence) explain your answer. (* FILL IN HERE *) *) (** [] *) End Temp5. End Temp4. (* ########################################################### *) (** * Multi-step reduction *) (** Until now, we've been working with the _single-step reduction relation_ [step], which formalizes the individual steps of _abstract machine_ for executing programs. It is also interesting to use this machine to reduce programs to completion, to find out what final result they yield. This can be formalized in two steps. - First, we define a _multi-step reduction relation_ [stepmany], which relates terms [t] and [t'] if [t] can reach [t'] by any number (including 0) of single reduction steps. - Then we can define a "result" of a term [t] as a normal form that [t] can reach by some number of reduction steps. Formally, we write [normal_form_of t t'] to mean that [t'] is a normal form reachable from [t] by many-step reduction. *) (* ########################################################### *) (** ** Reflexive, Transitive Closure *) (** To begin, let's review a bit of terminology from the basic theory of relations, which you probably saw at some point in a discrete math course. *) (** The _reflexive, transitive closure_ of a relation [R] is the smallest relation that contains [R] and that is both reflexive and transitive. Formally, it is defined like this in the Relations module of the Coq standard library: *) Module Excerpt_From_Coq_Relations_Library. Inductive clos_refl_trans {A: Type} {R: relation A} : relation A := | rt_step : forall x y, R x y -> clos_refl_trans x y | rt_refl : forall x, clos_refl_trans x x | rt_trans : forall x y z, clos_refl_trans x y -> clos_refl_trans y z -> clos_refl_trans x z. End Excerpt_From_Coq_Relations_Library. Implicit Arguments clos_refl_trans [A]. Tactic Notation "rt_cases" tactic(first) tactic(c) := first; [ c "rt_step" | c "rt_refl" | c "rt_trans" ]. (** For example, the reflexive and transitive closure of the [next_nat] relation coincides with the [le] relation. *) Inductive next_nat (n:nat) : nat -> Prop := | nn : next_nat n (S n). Theorem next_nat_closure_is_le : forall n m, (n <= m) <-> ((clos_refl_trans next_nat) n m). Proof. intros n m. split. Case "->". intro H. induction H. apply rt_refl. apply rt_trans with m. apply IHle. apply rt_step. apply nn. Case "<-". intro H. rt_cases (induction H) SCase. SCase "rt_step". inversion H. apply le_S. apply le_n. SCase "rt_refl". apply le_n. SCase "rt_trans". apply le_trans with y. apply IHclos_refl_trans1. apply IHclos_refl_trans2. Qed. (** The above definition of reflexive, transitive closure is natural -- it says, explicitly, that the reflexive and transitive closure of [R] is the least relation that includes [R] and that is closed under rules of reflexivity and transitivity. But it turns out that this definition is not very convenient for doing proofs -- the "nondeterminism" of the rt_trans rule can sometimes lead to tricky inductions. Here is a more useful definition... *) Inductive refl_step_closure (X:Type) (R: relation X) : X -> X -> Prop := | rsc_refl : forall (x : X), refl_step_closure X R x x | rsc_step : forall (x y z : X), R x y -> refl_step_closure X R y z -> refl_step_closure X R x z. Implicit Arguments refl_step_closure [X]. (** This new definition "bundles together" the [rtc_R] and [rtc_trans] rules into the single rule step. The left-hand premise of this step is a single use of [R], leading to a much simpler induction principle. Before we go on, we should check that the 2 definitions do indeed define the same relation... First, we prove two lemmas showing that [rsc] mimics the behavior of the two "missing " [rtc] constructors. *) Tactic Notation "rsc_cases" tactic(first) tactic(c) := first; [ c "rsc_refl" | c "rsc_step" ]. Theorem rsc_R : forall (X:Type) (R:relation X) (x y : X), R x y -> refl_step_closure R x y. Proof. intros X R x y r. apply rsc_step with y. apply r. apply rsc_refl. Qed. (** **** Exercise: 2 stars (rsc_trans) *) Theorem rsc_trans : forall (X:Type) (R: relation X) (x y z : X), refl_step_closure R x y -> refl_step_closure R y z -> refl_step_closure R x z. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars (rtc_rsc_coincide) *) Theorem rtc_rsc_coincide : forall (X:Type) (R: relation X) (x y : X), clos_refl_trans R x y <-> refl_step_closure R x y. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ########################################################### *) (** ** Multi-step reduction *) (** Now we're ready to define the _multi-step reduction relation_. *) Notation stepmany := (refl_step_closure step). (** (Note that we use [Notation] instead of [Definition] here. This means that [stepmany] will be automatically unfolded by Coq, which will simplify some of the proof automation later on.) *) (** A few examples... *) Lemma test_stepmany_1: stepmany (tm_plus (tm_plus (tm_const 0) (tm_const 3)) (tm_plus (tm_const 2) (tm_const 4))) (tm_const (plus (plus 0 3) (plus 2 4))). Proof. apply rsc_step with (tm_plus (tm_const (plus 0 3)) (tm_plus (tm_const 2) (tm_const 4))). apply ST_Plus1. apply ST_PlusConstConst. apply rsc_step with (tm_plus (tm_const (plus 0 3)) (tm_const (plus 2 4))). apply ST_Plus2. apply v_const. apply ST_PlusConstConst. apply rsc_R. apply ST_PlusConstConst. Qed. (** Here's another proof for the same example that uses [eapply] to avoid explicitly constructing all the intermediate terms. *) Lemma test_stepmany_1': stepmany (tm_plus (tm_plus (tm_const 0) (tm_const 3)) (tm_plus (tm_const 2) (tm_const 4))) (tm_const (plus (plus 0 3) (plus 2 4))). Proof. eapply rsc_step. apply ST_Plus1. apply ST_PlusConstConst. eapply rsc_step. apply ST_Plus2. split. apply ST_PlusConstConst. eapply rsc_step. apply ST_PlusConstConst. apply rsc_refl. Qed. (** **** Exercise: 1 star (test_stepmany_2) *) Lemma test_stepmany_2: stepmany (tm_const 3) (tm_const 3). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 1 star (test_stepmany_3) *) Lemma test_stepmany_3: stepmany (tm_plus (tm_const 0) (tm_const 3)) (tm_plus (tm_const 0) (tm_const 3)). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 2 stars (test_stepmany_4) *) Lemma test_stepmany_4: stepmany (tm_plus (tm_const 0) (tm_plus (tm_const 2) (tm_plus (tm_const 0) (tm_const 3)))) (tm_plus (tm_const 0) (tm_const (plus 2 (plus 0 3)))). Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ########################################################### *) (** ** Normal forms *) (** Now we define normal forms and relate them to values. *) Notation step_normal_form := (normal_form step). (** If [t] steps to [t'] in zero or more steps and [t'] is a normal form, we say that "[t'] is a normal form of [t]." *) Definition normal_form_of (t t' : tm) := (stepmany t t' /\ step_normal_form t'). (** We have already seen that single-step reduction is deterministic -- i.e., a given term can take a single step in at most one way. It follows from this that, if [t] can reach a normal form, then this normal form is unique -- i.e., [normal_form_of] is a partial function. In other words, we can actually pronounce [normal_form t t'] as "[t'] is _the_ normal form of [t]." *) (** **** Exercise: 3 stars, optional (test_stepmany_3) *) Theorem normal_forms_unique: partial_function normal_form_of. Proof. unfold partial_function. unfold normal_form_of. intros x y1 y2 P1 P2. destruct P1 as [P11 P12]. destruct P2 as [P21 P22]. generalize dependent y2. (* We recommend using this initial setup as-is! *) (* FILL IN HERE *) Admitted. (** [] *) (** Indeed, something stronger is true for this language (though not for all programming languages!): the reduction of ANY term [t] will eventually reach a normal form -- i.e., [normal_form_of] is a _total_ function. Formally, we say the [step] relation is _normalizing_. *) Definition normalizing (X:Type) (R:relation X) := forall t, exists t', (refl_step_closure R) t t' /\ normal_form R t'. Implicit Arguments normalizing [X]. (** To prove that [step] is normalizing, we need a few lemmas. First, we observe that, if [t] reduces to [t'] in many steps, then the same sequence of reduction steps is possible when [t] appears as the left-hand child of a [tm_plus] node, and similarly when [t] appears as the right-hand child of a [tm_plus] node whose left-hand child is a value. *) Lemma stepmany_congr_1 : forall t1 t1' t2, stepmany t1 t1' -> stepmany (tm_plus t1 t2) (tm_plus t1' t2). Proof. intros t1 t1' t2 H. rsc_cases (induction H) Case. Case "rsc_refl". apply rsc_refl. Case "rsc_step". apply rsc_step with (tm_plus y t2). apply ST_Plus1. apply H. apply IHrefl_step_closure. Qed. (** **** Exercise: 2 stars *) Lemma stepmany_congr_2 : forall t1 t2 t2', value t1 -> stepmany t2 t2' -> stepmany (tm_plus t1 t2) (tm_plus t1 t2'). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** Before proving it formally, let's sketch the proof that [step] is normalizing. Theorem: [[ forall t, exists t', stepmany t t' /\ normal_form step t' ]] Proof sketch: By induction on terms. There are two cases to consider: - [t = tm_const n] for some [n]. Here [t] doesn't take a step, and we have [t' = t]. We can derive the left-hand side by reflexivity and the right-hand side by observing (a) that values are normal forms (by [value_is_nf]) and (b) that t is a value (by [v_const]). - [t = tm_plus t1 t2] for some [t1] and [t2]. By the IH, both [t1] and [t2] normalize to normal forms [t1'] and [t2'], respectively. Recall that normal forms are values (by [nf_is_value]); we know that [t1' = tm_const n1] and [t2' = tm_const n2], for some [n1] and [n2]. We can combine the stepmany derivations for [t1] and [t2] to prove that: [[ stepmany (tm_plus t1 t2) (tm_const (plus n1 n2)). ]] It is clear that our choice of [t' = tm_const (plus n1 n2)] is a value, which is in turn a normal form. *) Theorem step_normalizing : normalizing step. Proof. unfold normalizing. tm_cases (induction t) Case. Case "tm_const". exists (tm_const n). split. SCase "l". apply rsc_refl. SCase "r". apply value_is_nf. apply v_const. Case "tm_plus". destruct IHt1 as [t1' H1]. destruct IHt2 as [t2' H2]. destruct H1 as [H11 H12]. destruct H2 as [H21 H22]. apply nf_is_value in H12. apply nf_is_value in H22. inversion H12 as [n1]. inversion H22 as [n2]. rewrite <- H in H11. rewrite <- H0 in H21. exists (tm_const (plus n1 n2)). split. SCase "l". apply rsc_trans with (tm_plus (tm_const n1) t2). apply stepmany_congr_1. apply H11. apply rsc_trans with (tm_plus (tm_const n1) (tm_const n2)). apply stepmany_congr_2. apply v_const. apply H21. apply rsc_R. apply ST_PlusConstConst. SCase "r". apply value_is_nf. apply v_const. Qed. (* ########################################################### *) (** ** Equivalence of big-step and small-step reduction *) (** Having defined the operational semantics of our tiny programming language in two different styles, it makes sense to ask whether these definitions actually define the same thing! They do, but it is not completely straightforward to show this, or even to see how to state it exactly, since one of the relations only goes a small step at a time while the other proceeds in large chunks. *) Lemma eval__value : forall t1 t2, eval t1 t2 -> value t2. Proof. intros t1 t2 HE. (eval_cases (inversion HE) Case); apply v_const. Qed. (** **** Exercise: 3 stars (eval__stepmany) *) (** You'll want to use the congruences and some properties of [rsc]. *) Theorem eval__stepmany : forall t v, eval t v -> stepmany t v. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars (eval__stepmany_inf) *) (** Write an informal version of the proof of eval__stepmany. (* FILL IN HERE *) [] *) (** **** Exercise: 3 stars (step__eval) *) Theorem step__eval : forall t t' v, step t t' -> eval t' v -> eval t v. Proof. (* FILL IN HERE *) Admitted. (** [] *) Theorem stepmany__eval : forall t v, normal_form_of t v -> eval t v. Proof. intros t v Hnorm. unfold normal_form_of in Hnorm. inversion Hnorm as [Hs Hnf]; clear Hnorm. (* v is a normal form -> v = tm_const n for some n *) apply nf_is_value in Hnf. inversion Hnf. clear Hnf. (rsc_cases (induction Hs) Case); subst. Case "rsc_refl". apply E_Const. Case "rsc_step". eapply step__eval. eassumption. apply IHHs. reflexivity. Qed. (** Bringing it all together into a crisp connection, we can simply say that the [v] is the normal form of [t] iff [t] evaluates to [v]. *) Corollary stepmany_iff_eval : forall t v, normal_form_of t v <-> eval t v. Proof. split. Case "->". apply stepmany__eval. Case "<-". unfold normal_form_of. intros E. split. apply eval__stepmany. assumption. apply value_is_nf. eapply eval__value. eassumption. Qed. (* ########################################################### *) (** ** Additional exercises *) (** **** Exercise: 4 stars, optional (interp_tm) *) (** Define a [Fixpoint] that evaluates [tm]s. Prove that it is equivalent to the existing semantics. Hint: we just proved that [eval] and [stepmany] are equivalent, so logically it doesn't matter which you choose. One will be easier than the other, though! *) (* FILL IN HERE *) (** [] *) (** **** Exercise: 4 stars, optional (combined_properties) *) (** We've considered the arithmetic and conditional expressions separately. This exercise explores how the two interact. *) Module Combined. Inductive tm : Type := | tm_const : nat -> tm | tm_plus : tm -> tm -> tm | tm_true : tm | tm_false : tm | tm_if : tm -> tm -> tm -> tm. Tactic Notation "tm_cases" tactic(first) tactic(c) := first; [ c "tm_const" | c "tm_plus" | c "tm_true" | c "tm_false" | c "tm_if" ]. Inductive value : tm -> Prop := | v_const : forall n, value (tm_const n) | v_true : value tm_true | v_false : value tm_false. Inductive step : tm -> tm -> Prop := | ST_PlusConstConst : forall n1 n2, step (tm_plus (tm_const n1) (tm_const n2)) (tm_const (plus n1 n2)) | ST_Plus1 : forall t1 t1' t2, (step t1 t1') -> step (tm_plus t1 t2) (tm_plus t1' t2) | ST_Plus2 : forall n1 t2 t2', (step t2 t2') -> step (tm_plus (tm_const n1) t2) (tm_plus (tm_const n1) t2') | ST_IfTrue : forall t1 t2, step (tm_if tm_true t1 t2) t1 | ST_IfFalse : forall t1 t2, step (tm_if tm_false t1 t2) t2 | ST_If : forall t1 t1' t2 t3, step t1 t1' -> step (tm_if t1 t2 t3) (tm_if t1' t2 t3). Tactic Notation "step_cases" tactic(first) tactic(c) := first; [ c "ST_PlusConstConst" | c "ST_Plus1" | c "ST_Plus2" | c "ST_IfTrue" | c "ST_IfFalse" | c "ST_If" ]. (** Earlier, we separately proved for both plus- and if-expressions - that the step relation was a partial function (i.e., it was deterministic), and - a progress lemma, stating that every term is either a value or can take step. Prove or disprove these for the combined langauge. *) (* FILL IN HERE *) (** [] *) End Combined.