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Proof step 10 from proof ac-in-10 (previous step)

is_fskey(sk1), r -- sk ∈ addclass(sk1 .fs .class, (sk1 × val) ' + fis0, st), r ∈ addclass(sk1 .fs .class, (sk1 × val) ' + fis0, st), ∀ sk0, class0, r0, st0. ¬ sk0 ∈ fis0 ∧ sk0 ≠ mkfs(class0, void_type, `initstate' .field) ' → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ sk = sk1, sk = mkfs(sk1 .fs .class, void_type, `initstate' .field) ', r -- sk ∈ st, sk ∈ fis0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
sk1 = fs ', is_fskey(sk1), r -- sk ∈ addclass(fs .class, (sk1 × val) ' + fis0, st), r ∈ addclass(fs .class, (sk1 × val) ' + fis0, st), ∀ sk0, class0, r0, st0. ¬ sk0 ∈ fis0 ∧ sk0 ≠ mkfs(class0, void_type, `initstate' .field) ' → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ sk = sk1, sk = mkfs(fs .class, void_type, `initstate' .field) ', r -- sk ∈ st, sk ∈ fis0

Rule argument

[speclemmaarg: ,is_fskey(sk) → (fs = sk .fs ↔ sk = fs ') ,[sk, fs] → [sk1, fs] ,T ,storekey , ,elim-03 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_fskey(sk) → (fs = sk .fs ↔ sk = fs ')])], [EXTRATERMS : (sk1, fs)]]