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Yixuan Chen
Iris
Commits
0c801b09
Commit
0c801b09
authored
9 years ago
by
Ralf Jung
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program_logic/invariants.v
+7
-6
7 additions, 6 deletions
program_logic/invariants.v
program_logic/pviewshifts.v
+5
-0
5 additions, 0 deletions
program_logic/pviewshifts.v
with
12 additions
and
6 deletions
program_logic/invariants.v
+
7
−
6
View file @
0c801b09
...
...
@@ -64,12 +64,13 @@ Global Instance inv_always_stable N P : AlwaysStable (inv N P) := _.
Lemma
always_inv
N
P
:
(
□
inv
N
P
)
%
I
≡
inv
N
P
.
Proof
.
by
rewrite
always_always
.
Qed
.
(* We actually pretty much lose the abolity to deal with mask-changing view
shifts when using `inv`. This is because we cannot exactly name the invariants
any more. But that's okay; all this means is that sugar like the atomic
triples will have to prove its own version of the open_close rule
by unfolding `inv`. *)
(* TODO Can we prove something that helps for both open_close lemmas? *)
(* There is not really a way to provide versions of pvs_openI and pvs_closeI
that work with inv. The issue is that these rules track the exact current
mask too precisely. However, we *can* provide abstract rules by
performing both the opening and the closing of the invariant in the rule,
and then implicitly framing all the unused invariants around the
"inner" view shift provided by the client. *)
Lemma
pvs_open_close
E
N
P
Q
:
nclose
N
⊆
E
→
(
inv
N
P
∧
(
▷
P
-★
pvs
(
E
∖
nclose
N
)
(
E
∖
nclose
N
)
(
▷
P
★
Q
)))
⊑
pvs
E
E
Q
.
...
...
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program_logic/pviewshifts.v
+
5
−
0
View file @
0c801b09
...
...
@@ -158,6 +158,11 @@ Lemma pvs_mask_frame_mono E1 E1' E2 E2' P Q :
P
⊑
Q
→
pvs
E1'
E2'
P
⊑
pvs
E1
E2
Q
.
Proof
.
intros
HE1
HE2
HEE
->
.
by
apply
pvs_mask_frame'
.
Qed
.
(* It should be possible to give a stronger version of this rule
that does not force the conclusion view shift to have twice the
same mask. However, even expressing the side-conditions on the
mask becomes really ugly then, and we have now found an instance
where that would be useful. *)
Lemma
pvs_trans3
E1
E2
Q
:
E2
⊆
E1
→
pvs
E1
E2
(
pvs
E2
E2
(
pvs
E2
E1
Q
))
⊑
pvs
E1
E1
Q
.
Proof
.
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