pl/frap
1
0
Fork 0
mirror of https://github.com/achlipala/frap.git synced 2024-09-19 19:57:13 +00:00
Code Issues Projects Releases Packages Wiki Activity Actions

Tweak OperationalSemantics_template.v

Browse source
This commit is contained in:
Adam Chlipala 2017-03-13 10:51:40 -04:00
parent c8322773a4
commit 0e06d062d1
1 changed files with 4 additions and 70 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

74
OperationalSemantics_template.v
View file

@ -367,7 +367,7 @@ Theorem big_small : forall v c v', eval v c v'
Proof.
Admitted.
Theorem small_big_snazzy : forall v c v', step^* (v, c) (v', Skip)
Theorem small_big : forall v c v', step^* (v, c) (v', Skip)
-> eval v c v'.
Proof.
Admitted.
@ -741,33 +741,8 @@ Module Concurrent.
Proof.
eexists; propositional.
unfold prog.
Admitted.
econstructor.
eapply CStep with (C := CPar1 (CSeq Hole _) _); eauto.
econstructor.
eapply CStep with (C := CPar2 _ (CSeq Hole _)); eauto.
econstructor.
eapply CStep with (C := CPar1 Hole _); eauto.
econstructor.
eapply CStep with (C := CPar2 _ Hole); eauto.
econstructor.
eapply CStep with (C := CPar1 Hole _); eauto.
econstructor.
eapply CStep with (C := Hole); eauto.
econstructor.
eapply CStep with (C := Hole); eauto.
econstructor.
simplify.
equality.
Qed.
(** Proving equivalence with non-contextual semantics *)
@ -806,49 +781,8 @@ Module Concurrent.
Hint Constructors step.
(* Now, an automated proof of equivalence. Actually, it's *exactly* the same
* proof we used for the old feature set! *)
Theorem step_cstep : forall v c v' c',
step (v, c) (v', c')
-> cstep (v, c) (v', c').
Proof.
induct 1; repeat match goal with
| [ H : forall a b c d, _ = _ -> _ = _ -> _ |- _ ] =>
specialize (H _ _ _ _ eq_refl eq_refl)
| [ H : cstep _ _ |- _ ] => invert H
end; eauto.
Qed.
Hint Resolve step_cstep.
Lemma step0_step : forall v c v' c',
step0 (v, c) (v', c')
-> step (v, c) (v', c').
Proof.
induct 1; eauto.
Qed.
Hint Resolve step0_step.
Lemma cstep_step' : forall C c0 c,
plug C c0 c
-> forall v' c'0 v c', step0 (v, c0) (v', c'0)
-> plug C c'0 c'
-> step (v, c) (v', c').
Proof.
induct 1; simplify; repeat match goal with
| [ H : plug _ _ _ |- _ ] => invert1 H
end; eauto.
Qed.
Hint Resolve cstep_step'.
Theorem cstep_step : forall v c v' c',
cstep (v, c) (v', c')
-> step (v, c) (v', c').
Proof.
induct 1; eauto.
Qed.
* proof we used for the old feature set! For full dramatic effect, copy and
* paste here from above. *)
End Concurrent.