# Theory Big_Step

theory Big_Step
imports Com
(* 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 parameter 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: "⟦ (c1,s1) ⇒ s2;  (c2,s2) ⇒ s3 ⟧ ⟹ (c1;;c2, s1) ⇒ s3" |
IfTrue: "⟦ bval b s;  (c1,s) ⇒ t ⟧ ⟹ (IF b THEN c1 ELSE c2, s) ⇒ t" |
IfFalse: "⟦ ¬bval b s;  (c2,s) ⇒ t ⟧ ⟹ (IF b THEN c1 ELSE c2, s) ⇒ t" |
WhileFalse: "¬bval b s ⟹ (WHILE b DO c,s) ⇒ s" |
WhileTrue:
"⟦ bval b s1;  (c,s1) ⇒ s2;  (WHILE b DO c, s2) ⇒ s3 ⟧
⟹ (WHILE b DO c, s1) ⇒ s3"
text_raw{*}%endsnip*}

text_raw{*\snip{BigStepEx}{1}{2}{% *}
schematic_goal 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

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 s1 t t'
-- "The assumptions of the rule:"
assume "bval b s" and "(c,s) ⇒ s1" and "(WHILE b DO c,s1) ⇒ t"
-- {* Ind.Hyp; note the @{text"⋀"} because of arbitrary: *}
assume IHc: "⋀t'. (c,s) ⇒ t' ⟹ t' = s1"
assume IHw: "⋀t'. (WHILE b DO c,s1) ⇒ t' ⟹ t' = t"
-- "Premise of implication:"
assume "(WHILE b DO c,s) ⇒ t'"
with bval b s obtain s1' where
c: "(c,s) ⇒ s1'" and
w: "(WHILE b DO c,s1') ⇒ t'"
by auto
from c IHc have "s1' = s1" by blast
with w IHw show "t' = t" by blast
qed blast+ -- "prove the rest automatically"
text_raw{*}%endsnip*}

end