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Proof step 23 from proof ac-lookup-05 (previous step)

¬ is_indexkey((sk0 × val) .1), is_indexkey(sk), ∀ sk0, class0, st0, r0. ¬ is_fskey(sk0) → (addclass(class0, fis0, st0)[r0 -- sk0]) = st0[r0 -- sk0] ⊦ (addclass(class, (sk0 × val) ' + fis0, st)[r -- sk]) = st[r -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
is_indexkey(sk), is_fskey(sk0), ∀ sk0, class0, st0, r0. ¬ is_fskey(sk0) → (addclass(class0, fis0, st0)[r0 -- sk0]) = st0[r0 -- sk0] ⊦ (addclass(class, (sk0 × val) ' + fis0, st)[r -- sk]) = st[r -- sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ is_indexkey(sk) ↔ is_fskey(sk)], [CSIMPSEQ : ⊦ (sk × val) .1 = sk])]]