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Proof step 10 from proof ac-in-11 (previous step)

is_fskey(sk1), is_indexkey(sk), r -- sk ∈ addclass(sk1 .fs .class, (sk1 × val) ' + fis0, st), r ∈ addclass(sk1 .fs .class, (sk1 × val) ' + fis0, st), ∀ sk0, r0, class0, st0. ¬ is_fskey(sk0) → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ r -- sk ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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)]]