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Proof step 10 from proof ac-lookup-06 (previous step)

¬ (is_fskey((sk0 × val) .1) ∧ (sk0 × val) .1 .fs .class ≠ class), is_fskey(sk0), ∀ sk, class, st, r. sk = mkfs(class, void_type, `initstate' .field) ' ∨ (addclass(class, fis0, st)[r -- sk]) = st[r -- sk] ∨ sk ∈ fis0 ⊦ sk = mkfs(class, void_type, `initstate' .field) ', (addclass(class, (sk0 × val) ' + fis0, st)[r -- sk]) = st[r -- sk], sk = sk0, sk ∈ fis0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
is_fskey(sk0), ∀ sk, class, st, r. sk = mkfs(class, void_type, `initstate' .field) ' ∨ (addclass(class, fis0, st)[r -- sk]) = st[r -- sk] ∨ sk ∈ fis0 ⊦ sk = sk0, (addclass(sk0 .fs .class, (sk0 × val) ' + fis0, st)[r -- sk]) = st[r -- sk], sk = mkfs(sk0 .fs .class, void_type, `initstate' .field) ', sk ∈ fis0

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])]]