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Proof step 4 from proof ac-no1 (previous step)

is_fskey(sk) ⊦ sk .fs .class = class, addclass(class, (sk × val) ' + fis, st) = addclass(class, fis, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
sk = fs ', is_fskey(sk) ⊦ fs .class = class, addclass(class, (sk × val) ' + fis, st) = addclass(class, fis, st)

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)]]