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Proof step 10 from proof ac-in-02 (previous step)

is_fskey(sk), r -- i ' ∈ addclass(sk .fs .class, (sk × val) ' + fis0, st), r ∈ addclass(sk .fs .class, (sk × val) ' + fis0, st), ∀ r0, i0, class, st0. r0 -- i0 ' ∈ addclass(class, fis0, st0) → r0 -- i0 ' ∈ st0 ⊦ r -- i ' ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
sk = fs ', is_fskey(sk), r -- i ' ∈ addclass(fs .class, (sk × val) ' + fis0, st), r ∈ addclass(fs .class, (sk × val) ' + fis0, st), ∀ r0, i0, class, st0. r0 -- i0 ' ∈ addclass(class, fis0, st0) → r0 -- i0 ' ∈ st0 ⊦ r -- i ' ∈ 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)]]