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

¬ (is_fskey((sk1 × val) .1) ∧ (sk1 × val) .1 .fs .class ≠ class), r -- sk ∈ addclass(class, (sk1 × val) ' + fis0, st), is_indexkey(sk), is_fskey(sk1), r ∈ addclass(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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [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])]]