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Proof step 15 from proof ac-lookup-05 (previous step)

¬ (is_fskey((sk0 × val) .1) ∧ (sk0 × val) .1 .fs .class ≠ class), 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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
is_fskey(sk0), is_indexkey(sk), ∀ sk0, class0, st0, r0. ¬ is_fskey(sk0) → (addclass(class0, fis0, st0)[r0 -- sk0]) = st0[r0 -- sk0] ⊦ (addclass(sk0 .fs .class, (sk0 × val) ' + fis0, st)[r -- sk]) = st[r -- sk]

Rule argument

[emptyarg]

Additional information

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