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Proof step 19 from proof ac-in-09 (previous step)

is_fskey(sk), ∀ r, class1, st. r = jvmref ∨ (r ∈ addclass(class1, fis0, st) ↔ r ∈ st) ⊦ r = jvmref, r ∈ addclass(class1, fis0, st) ↔ r ∈ st, sk .fs .class = class1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
sk = fs ', is_fskey(sk), ∀ r, class1, st. r = jvmref ∨ (r ∈ addclass(class1, fis0, st) ↔ r ∈ st) ⊦ r = jvmref, r ∈ addclass(class1, fis0, st) ↔ r ∈ st, fs .class = class1

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)]]