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Proof step 8 from proof ac-dups-05 (previous step)

¬ (is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class1), is_fskey(sk) ⊦ addclass(class1, (sk × val) ' + fis1 + (sk × val0) ', st) = addclass(class1, (sk × val) ' + fis1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
is_fskey(sk) ⊦ addclass(class1, (sk × val) ' + fis1 + (sk × val0) ', st) = addclass(class1, (sk × val) ' + fis1, 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])]]