(** Formal Reasoning About Programs * Chapter 3: Semantics via Interpreters * Author: Adam Chlipala * License: https://creativecommons.org/licenses/by-nc-nd/4.0/ *) Require Import Frap. (* We begin with a return to our arithmetic language from the last chapter, * adding subtraction, which will come in handy later. *) Inductive arith : Set := | Const (n : nat) | Var (x : var) | Plus (e1 e2 : arith) | Minus (e1 e2 : arith) | Times (e1 e2 : arith). Example ex1 := Const 42. Example ex2 := Plus (Var "y") (Times (Var "x") (Const 3)). (* The above definition only explains what programs *look like*. * We also care about what they *mean*. * The natural meaning of an expression is the number it evaluates to. * Actually, it's not quite that simple. * We need to consider the meaning to be a function over a valuation * to the variables, which in turn is itself a finite map from variable * names to numbers. We use the book library's [map] type family. *) Definition valuation := map var nat. Fixpoint interp (e : arith) (v : valuation) : nat := match e with | Const n => n | Var x => (* Note use of infix operator to look up a key in a finite map. *) match v $? x with | None => 0 (* goofy default value! *) | Some n => n end | Plus e1 e2 => interp e1 v + interp e2 v | Minus e1 e2 => interp e1 v - interp e2 v (* For anyone who's wondering: this [-] sticks at 0, * if we would otherwise underflow. *) | Times e1 e2 => interp e1 v * interp e2 v end. (* Here's an example valuation, using an infix operator for map extension. *) Definition valuation0 : valuation := $0 $+ ("x", 17) $+ ("y", 3). (* Unfortunately, we can't execute code based on finite maps, since, for * convenience, they use uncomputable features. The reason is that we need a * comparison function, a hash function, etc., to do computable finite-map * implementation, and such things are impossible to compute automatically for * all types in Coq. However, we can still prove theorems about execution of * finite-map programs, and the [simplify] tactics knows how to reduce the * key constructions. *) Theorem interp_ex1 : interp ex1 valuation0 = 42. Proof. simplify. equality. Qed. Theorem interp_ex2 : interp ex2 valuation0 = 54. Proof. unfold valuation0. simplify. equality. Qed. (* Here's the silly transformation we defined last time. *) Fixpoint commuter (e : arith) : arith := match e with | Const _ => e | Var _ => e | Plus e1 e2 => Plus (commuter e2) (commuter e1) | Minus e1 e2 => Minus (commuter e1) (commuter e2) | Times e1 e2 => Times (commuter e2) (commuter e1) end. (* Instead of proving various odds-and-ends properties about it, * let's show what we *really* care about: it preserves the * *meanings* of expressions! *) Theorem commuter_ok : forall v e, interp (commuter e) v = interp e v. Proof. induct e; simplify. equality. equality. linear_arithmetic. equality. rewrite IHe1. rewrite IHe2. ring. Qed. (* Well, that's a relief! ;-) *) (* Let's also revisit substitution. *) Fixpoint substitute (inThis : arith) (replaceThis : var) (withThis : arith) : arith := match inThis with | Const _ => inThis | Var x => if x ==v replaceThis then withThis else inThis | Plus e1 e2 => Plus (substitute e1 replaceThis withThis) (substitute e2 replaceThis withThis) | Minus e1 e2 => Minus (substitute e1 replaceThis withThis) (substitute e2 replaceThis withThis) | Times e1 e2 => Times (substitute e1 replaceThis withThis) (substitute e2 replaceThis withThis) end. (* Note the use of an infix operator for overriding one entry in a finite * map. *) Theorem substitute_ok : forall v replaceThis withThis inThis, interp (substitute inThis replaceThis withThis) v = interp inThis (v $+ (replaceThis, interp withThis v)). Proof. induct inThis; simplify; try equality. (* One case left after our basic heuristic: * the variable case, naturally! *) cases (x ==v replaceThis); simplify; try equality. Qed. (* Great; we seem to have gotten that one right, too. *) (* Let's also defined a pared-down version of the expression-simplificaton * functions from last chapter. *) Fixpoint doSomeArithmetic (e : arith) : arith := match e with | Const _ => e | Var _ => e | Plus (Const n1) (Const n2) => Const (n1 + n2) | Plus e1 e2 => Plus (doSomeArithmetic e1) (doSomeArithmetic e2) | Minus e1 e2 => Minus (doSomeArithmetic e1) (doSomeArithmetic e2) | Times (Const n1) (Const n2) => Const (n1 * n2) | Times e1 e2 => Times (doSomeArithmetic e1) (doSomeArithmetic e2) end. Theorem doSomeArithmetic_ok : forall e v, interp (doSomeArithmetic e) v = interp e v. Proof. induct e; simplify; try equality. cases e1; simplify; try equality. cases e2; simplify; equality. cases e1; simplify; try equality. cases e2; simplify; equality. Qed. (* Of course, we're going to get bored if we confine ourselves to arithmetic * expressions for the rest of our journey. Let's get a bit fancier and define * a *stack machine*, related to postfix calculators that some of you may have * experienced. *) Inductive instruction := | PushConst (n : nat) | PushVar (x : var) | Add | Subtract | Multiply. (* What does it all mean? An interpreter tells us unambiguously! *) Definition run1 (i : instruction) (v : valuation) (stack : list nat) : list nat := match i with | PushConst n => n :: stack | PushVar x => (match v $? x with | None => 0 | Some n => n end) :: stack | Add => match stack with | arg2 :: arg1 :: stack' => arg1 + arg2 :: stack' | _ => stack (* arbitrary behavior in erroneous case *) end | Subtract => match stack with | arg2 :: arg1 :: stack' => arg1 - arg2 :: stack' | _ => stack (* arbitrary behavior in erroneous case *) end | Multiply => match stack with | arg2 :: arg1 :: stack' => arg1 * arg2 :: stack' | _ => stack (* arbitrary behavior in erroneous case *) end end. (* That function explained how to run one instruction. * Here's how to run several of them. *) Fixpoint run (is : list instruction) (v : valuation) (stack : list nat) : list nat := match is with | nil => stack | i :: is' => run is' v (run1 i v stack) end. (* Instead of writing fiddly stack programs ourselves, let's *compile* * arithmetic expressions into equivalent stack programs. *) Fixpoint compile (e : arith) : list instruction := match e with | Const n => PushConst n :: nil | Var x => PushVar x :: nil | Plus e1 e2 => compile e1 ++ compile e2 ++ Add :: nil | Minus e1 e2 => compile e1 ++ compile e2 ++ Subtract :: nil | Times e1 e2 => compile e1 ++ compile e2 ++ Multiply :: nil end. (* Now, of course, we should prove our compiler correct. * Skip down to the next theorem to see the overall correctness statement. * It turns out that we need to strengthen the induction hypothesis with a * lemma, to push the proof through. *) Lemma compile_ok' : forall e v is stack, run (compile e ++ is) v stack = run is v (interp e v :: stack). Proof. induct e; simplify. equality. equality. (* Here we want to use associativity of [++], to get the conclusion to match * an induction hypothesis. Let's ask Coq to search its library for lemmas * that would justify such a rewrite, giving a pattern with wildcards, to * specify the essential structure that the rewrite should match. *) SearchRewrite ((_ ++ _) ++ _). (* Ah, we see just the one! *) rewrite app_assoc_reverse. rewrite IHe1. rewrite app_assoc_reverse. rewrite IHe2. simplify. equality. rewrite app_assoc_reverse. rewrite IHe1. rewrite app_assoc_reverse. rewrite IHe2. simplify. equality. rewrite app_assoc_reverse. rewrite IHe1. rewrite app_assoc_reverse. rewrite IHe2. simplify. equality. Qed. (* The overall theorem follows as a simple corollary. *) Theorem compile_ok : forall e v, run (compile e) v nil = interp e v :: nil. Proof. simplify. (* To match the form of our lemma, we need to replace [compile e] with * [compile e ++ nil], adding a "pointless" concatenation of the empty list. * [SearchRewrite] again helps us find a library lemma. *) SearchRewrite (_ ++ nil). rewrite (app_nil_end (compile e)). (* Note that we can use [rewrite] with explicit values of the first few * quantified variables of a lemma. Otherwise, [rewrite] picks an * unhelpful place to rewrite. (Try it and see!) *) apply compile_ok'. (* Direct appeal to a previously proved lemma *) Qed. (* Let's get a bit fancier, moving toward the level of general-purpose * imperative languages. Here's a language of commands, building on the * language of expressions we have defined. *) Inductive cmd := | Skip | Assign (x : var) (e : arith) | Sequence (c1 c2 : cmd) | Repeat (e : arith) (body : cmd). (* That last constructor is for repeating a body command some number of * times. Note that we sneakily avoid constructs that could introduce * nontermination, since Coq only accepts terminating programs, and we want to * write an interpreter for commands. * In contrast to our last one, this interpreter *transforms valuations*. * We use a helper function for self-composing a function some number of * times. *) Fixpoint selfCompose {A} (f : A -> A) (n : nat) : A -> A := match n with | O => fun x => x | S n' => fun x => selfCompose f n' (f x) end. Fixpoint exec (c : cmd) (v : valuation) : valuation := match c with | Skip => v | Assign x e => v $+ (x, interp e v) | Sequence c1 c2 => exec c2 (exec c1 v) | Repeat e body => selfCompose (exec body) (interp e v) v end. (* Let's define some programs and prove that they operate in certain ways. *) Example factorial_ugly := Sequence (Assign "output" (Const 1)) (Repeat (Var "input") (Sequence (Assign "output" (Times (Var "output") (Var "input"))) (Assign "input" (Minus (Var "input") (Const 1))))). (* Ouch; that code is hard to read. Let's introduce some notations to make the * concrete syntax more palatable. We won't explain the general mechanisms on * display here, but see the Coq manual for details, or try to reverse-engineer * them from our examples. *) Coercion Const : nat >-> arith. Coercion Var : var >-> arith. Infix "+" := Plus : arith_scope. Infix "-" := Minus : arith_scope. Infix "*" := Times : arith_scope. Delimit Scope arith_scope with arith. Notation "x <- e" := (Assign x e%arith) (at level 75). Infix ";" := Sequence (at level 76). Notation "'repeat' e 'doing' body 'done'" := (Repeat e%arith body) (at level 75). (* OK, let's try that program again. *) Example factorial := "output" <- 1; repeat "input" doing "output" <- "output" * "input"; "input" <- "input" - 1 done. (* Now we prove that it really computes factorial. * First, a reference implementation as a functional program. *) Fixpoint fact (n : nat) : nat := match n with | O => 1 | S n' => n * fact n' end. (* To prove that [factorial] is correct, the real action is in a lemma, to be * proved by induction, showing that the loop works correctly. So, let's first * assign a name to the loop body alone. *) Definition factorial_body := "output" <- "output" * "input"; "input" <- "input" - 1. (* Now for that lemma: self-composition of the body's semantics produces the * expected changes in the valuation. * Note that here we're careful to put the quantified variable [input] *first*, * because the variables coming after it will need to *change* in the course of * the induction. Try switching the order to see what goes wrong if we put e * [input] later. *) Lemma factorial_ok' : forall input output v, v $? "input" = Some input -> v $? "output" = Some output -> selfCompose (exec factorial_body) input v = v $+ ("input", 0) $+ ("output", output * fact input). Proof. induct input; simplify. maps_equal. (* [maps_equal]: prove that two finite maps are equal by considering all * the relevant cases for mappings of different keys. *) rewrite H0. f_equal. linear_arithmetic. trivial. (* [trivial]: Coq maintains a database of simple proof steps, such as proving * a fact by direct appeal to a matching hypothesis. [trivial] asks to try * all such simple steps. *) rewrite H, H0. (* Note the two arguments to one [rewrite]! *) rewrite (IHinput (output * S input)). (* Note the careful choice of a quantifier instantiation for the IH! *) maps_equal. f_equal; ring. simplify; f_equal; linear_arithmetic. simplify; equality. Qed. (* Finally, we have the natural correctness condition for factorial as a whole * program. *) Theorem factorial_ok : forall v input, v $? "input" = Some input -> exec factorial v $? "output" = Some (fact input). Proof. simplify. rewrite H. rewrite (factorial_ok' input 1); simplify. f_equal; linear_arithmetic. trivial. trivial. Qed.