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Proof step 12 from proof ac-in-09 (previous step)

¬ (is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class1), is_fskey(sk), ∀ r, class1, st. r = jvmref ∨ (r ∈ addclass(class1, fis0, st) ↔ r ∈ st) ⊦ r = jvmref, r ∈ addclass(class1, (sk × val) ' + fis0, st) ↔ r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
is_fskey(sk), ∀ r, class1, st. r = jvmref ∨ (r ∈ addclass(class1, fis0, st) ↔ r ∈ st) ⊦ r = jvmref, r ∈ addclass(class1, (sk × val) ' + fis0, st) ↔ r ∈ st, ¬ is_indexkey(sk) ∧ sk .fs .class ≠ class1

Rule argument

[emptyarg]

Additional information

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