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Proof step 23 from proof ac-in-06 (previous step)

is_fskey(sk), ∀ class1, class2, ty, fieldvar, st. class1 = class2 ∨ (jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st) ⊦ class1 = class2, jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st, sk .fs .class = class2

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
sk = fs ', is_fskey(sk), ∀ class1, class2, ty, fieldvar, st. class1 = class2 ∨ (jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st) ⊦ class1 = class2, jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st, fs .class = class2

Rule argument

[speclemmaarg: ,is_fskey(sk) → (fs = sk .fs ↔ sk = fs ') ,[sk, fs] → [sk, fs] ,T ,storekey , ,elim-03 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_fskey(sk) → (fs = sk .fs ↔ sk = fs ')])], [EXTRATERMS : (sk, fs)]]