(* Author: Gerwin Klein, Tobias Nipkow *) theory Big_Step imports Com begin subsection "Big-Step Semantics of Commands" text {* The big-step semantics is a straight-forward inductive definition with concrete syntax. Note that the first paramenter is a tuple, so the syntax becomes @{text "(c,s) => s'"}. *} text_raw{*\snip{BigStepdef}{0}{1}{% *} inductive big_step :: "com × state => state => bool" (infix "=>" 55) where Skip: "(SKIP,s) => s" | Assign: "(x ::= a,s) => s(x := aval a s)" | Seq: "[| (c⇩_{1},s⇩_{1}) => s⇩_{2}; (c⇩_{2},s⇩_{2}) => s⇩_{3}|] ==> (c⇩_{1};;c⇩_{2}, s⇩_{1}) => s⇩_{3}" | IfTrue: "[| bval b s; (c⇩_{1},s) => t |] ==> (IF b THEN c⇩_{1}ELSE c⇩_{2}, s) => t" | IfFalse: "[| ¬bval b s; (c⇩_{2},s) => t |] ==> (IF b THEN c⇩_{1}ELSE c⇩_{2}, s) => t" | WhileFalse: "¬bval b s ==> (WHILE b DO c,s) => s" | WhileTrue: "[| bval b s⇩_{1}; (c,s⇩_{1}) => s⇩_{2}; (WHILE b DO c, s⇩_{2}) => s⇩_{3}|] ==> (WHILE b DO c, s⇩_{1}) => s⇩_{3}" text_raw{*}%endsnip*} text_raw{*\snip{BigStepEx}{1}{2}{% *} schematic_lemma ex: "(''x'' ::= N 5;; ''y'' ::= V ''x'', s) => ?t" apply(rule Seq) apply(rule Assign) apply simp apply(rule Assign) done text_raw{*}%endsnip*} thm ex[simplified] text{* We want to execute the big-step rules: *} code_pred big_step . text{* For inductive definitions we need command \texttt{values} instead of \texttt{value}. *} values "{t. (SKIP, λ_. 0) => t}" text{* We need to translate the result state into a list to display it. *} values "{map t [''x''] |t. (SKIP, <''x'' := 42>) => t}" values "{map t [''x''] |t. (''x'' ::= N 2, <''x'' := 42>) => t}" values "{map t [''x'',''y''] |t. (WHILE Less (V ''x'') (V ''y'') DO (''x'' ::= Plus (V ''x'') (N 5)), <''x'' := 0, ''y'' := 13>) => t}" text{* Proof automation: *} text {* The introduction rules are good for automatically construction small program executions. The recursive cases may require backtracking, so we declare the set as unsafe intro rules. *} declare big_step.intros [intro] text{* The standard induction rule @{thm [display] big_step.induct [no_vars]} *} thm big_step.induct text{* This induction schema is almost perfect for our purposes, but our trick for reusing the tuple syntax means that the induction schema has two parameters instead of the @{text c}, @{text s}, and @{text s'} that we are likely to encounter. Splitting the tuple parameter fixes this: *} lemmas big_step_induct = big_step.induct[split_format(complete)] thm big_step_induct text {* @{thm [display] big_step_induct [no_vars]} *} subsection "Rule inversion" text{* What can we deduce from @{prop "(SKIP,s) => t"} ? That @{prop "s = t"}. This is how we can automatically prove it: *} inductive_cases SkipE[elim!]: "(SKIP,s) => t" thm SkipE text{* This is an \emph{elimination rule}. The [elim] attribute tells auto, blast and friends (but not simp!) to use it automatically; [elim!] means that it is applied eagerly. Similarly for the other commands: *} inductive_cases AssignE[elim!]: "(x ::= a,s) => t" thm AssignE inductive_cases SeqE[elim!]: "(c1;;c2,s1) => s3" thm SeqE inductive_cases IfE[elim!]: "(IF b THEN c1 ELSE c2,s) => t" thm IfE inductive_cases WhileE[elim]: "(WHILE b DO c,s) => t" thm WhileE text{* Only [elim]: [elim!] would not terminate. *} text{* An automatic example: *} lemma "(IF b THEN SKIP ELSE SKIP, s) => t ==> t = s" by blast text{* Rule inversion by hand via the ``cases'' method: *} lemma assumes "(IF b THEN SKIP ELSE SKIP, s) => t" shows "t = s" proof- from assms show ?thesis proof cases --"inverting assms" case IfTrue thm IfTrue thus ?thesis by blast next case IfFalse thus ?thesis by blast qed qed (* Using rule inversion to prove simplification rules: *) lemma assign_simp: "(x ::= a,s) => s' <-> (s' = s(x := aval a s))" by auto text {* An example combining rule inversion and derivations *} lemma Seq_assoc: "(c1;; c2;; c3, s) => s' <-> (c1;; (c2;; c3), s) => s'" proof assume "(c1;; c2;; c3, s) => s'" then obtain s1 s2 where c1: "(c1, s) => s1" and c2: "(c2, s1) => s2" and c3: "(c3, s2) => s'" by auto from c2 c3 have "(c2;; c3, s1) => s'" by (rule Seq) with c1 show "(c1;; (c2;; c3), s) => s'" by (rule Seq) next -- "The other direction is analogous" assume "(c1;; (c2;; c3), s) => s'" thus "(c1;; c2;; c3, s) => s'" by auto qed subsection "Command Equivalence" text {* We call two statements @{text c} and @{text c'} equivalent wrt.\ the big-step semantics when \emph{@{text c} started in @{text s} terminates in @{text s'} iff @{text c'} started in the same @{text s} also terminates in the same @{text s'}}. Formally: *} text_raw{*\snip{BigStepEquiv}{0}{1}{% *} abbreviation equiv_c :: "com => com => bool" (infix "∼" 50) where "c ∼ c' ≡ (∀s t. (c,s) => t = (c',s) => t)" text_raw{*}%endsnip*} text {* Warning: @{text"∼"} is the symbol written \verb!\ < s i m >! (without spaces). As an example, we show that loop unfolding is an equivalence transformation on programs: *} lemma unfold_while: "(WHILE b DO c) ∼ (IF b THEN c;; WHILE b DO c ELSE SKIP)" (is "?w ∼ ?iw") proof - -- "to show the equivalence, we look at the derivation tree for" -- "each side and from that construct a derivation tree for the other side" { fix s t assume "(?w, s) => t" -- "as a first thing we note that, if @{text b} is @{text False} in state @{text s}," -- "then both statements do nothing:" { assume "¬bval b s" hence "t = s" using `(?w,s) => t` by blast hence "(?iw, s) => t" using `¬bval b s` by blast } moreover -- "on the other hand, if @{text b} is @{text True} in state @{text s}," -- {* then only the @{text WhileTrue} rule can have been used to derive @{text "(?w, s) => t"} *} { assume "bval b s" with `(?w, s) => t` obtain s' where "(c, s) => s'" and "(?w, s') => t" by auto -- "now we can build a derivation tree for the @{text IF}" -- "first, the body of the True-branch:" hence "(c;; ?w, s) => t" by (rule Seq) -- "then the whole @{text IF}" with `bval b s` have "(?iw, s) => t" by (rule IfTrue) } ultimately -- "both cases together give us what we want:" have "(?iw, s) => t" by blast } moreover -- "now the other direction:" { fix s t assume "(?iw, s) => t" -- "again, if @{text b} is @{text False} in state @{text s}, then the False-branch" -- "of the @{text IF} is executed, and both statements do nothing:" { assume "¬bval b s" hence "s = t" using `(?iw, s) => t` by blast hence "(?w, s) => t" using `¬bval b s` by blast } moreover -- "on the other hand, if @{text b} is @{text True} in state @{text s}," -- {* then this time only the @{text IfTrue} rule can have be used *} { assume "bval b s" with `(?iw, s) => t` have "(c;; ?w, s) => t" by auto -- "and for this, only the Seq-rule is applicable:" then obtain s' where "(c, s) => s'" and "(?w, s') => t" by auto -- "with this information, we can build a derivation tree for the @{text WHILE}" with `bval b s` have "(?w, s) => t" by (rule WhileTrue) } ultimately -- "both cases together again give us what we want:" have "(?w, s) => t" by blast } ultimately show ?thesis by blast qed text {* Luckily, such lengthy proofs are seldom necessary. Isabelle can prove many such facts automatically. *} lemma while_unfold: "(WHILE b DO c) ∼ (IF b THEN c;; WHILE b DO c ELSE SKIP)" by blast lemma triv_if: "(IF b THEN c ELSE c) ∼ c" by blast lemma commute_if: "(IF b1 THEN (IF b2 THEN c11 ELSE c12) ELSE c2) ∼ (IF b2 THEN (IF b1 THEN c11 ELSE c2) ELSE (IF b1 THEN c12 ELSE c2))" by blast lemma sim_while_cong_aux: "(WHILE b DO c,s) => t ==> c ∼ c' ==> (WHILE b DO c',s) => t" apply(induction "WHILE b DO c" s t arbitrary: b c rule: big_step_induct) apply blast apply blast done lemma sim_while_cong: "c ∼ c' ==> WHILE b DO c ∼ WHILE b DO c'" by (metis sim_while_cong_aux) text {* Command equivalence is an equivalence relation, i.e.\ it is reflexive, symmetric, and transitive. Because we used an abbreviation above, Isabelle derives this automatically. *} lemma sim_refl: "c ∼ c" by simp lemma sim_sym: "(c ∼ c') = (c' ∼ c)" by auto lemma sim_trans: "c ∼ c' ==> c' ∼ c'' ==> c ∼ c''" by auto subsection "Execution is deterministic" text {* This proof is automatic. *} theorem big_step_determ: "[| (c,s) => t; (c,s) => u |] ==> u = t" by (induction arbitrary: u rule: big_step.induct) blast+ text {* This is the proof as you might present it in a lecture. The remaining cases are simple enough to be proved automatically: *} text_raw{*\snip{BigStepDetLong}{0}{2}{% *} theorem "(c,s) => t ==> (c,s) => t' ==> t' = t" proof (induction arbitrary: t' rule: big_step.induct) -- "the only interesting case, @{text WhileTrue}:" fix b c s s⇩_{1}t t' -- "The assumptions of the rule:" assume "bval b s" and "(c,s) => s⇩_{1}" and "(WHILE b DO c,s⇩_{1}) => t" -- {* Ind.Hyp; note the @{text"!!"} because of arbitrary: *} assume IHc: "!!t'. (c,s) => t' ==> t' = s⇩_{1}" assume IHw: "!!t'. (WHILE b DO c,s⇩_{1}) => t' ==> t' = t" -- "Premise of implication:" assume "(WHILE b DO c,s) => t'" with `bval b s` obtain s⇩_{1}' where c: "(c,s) => s⇩_{1}'" and w: "(WHILE b DO c,s⇩_{1}') => t'" by auto from c IHc have "s⇩_{1}' = s⇩_{1}" by blast with w IHw show "t' = t" by blast qed blast+ -- "prove the rest automatically" text_raw{*}%endsnip*} end