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Proof step 11 from proof ac-bad-03 (previous step)

¬ (is_fskey((sk × val0) .1) ∧ (sk × val0) .1 .fs .class ≠ class1), is_fskey(sk), ∀ class1, i, val, st. addclass(class1, fis0 + (i ' × val) ', st) = addclass(class1, fis0, st) ⊦ addclass(class1, ((sk × val0) ' + fis0) + (i ' × val) ', st) = addclass(class1, (sk × val0) ' + fis0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
is_fskey(sk), ∀ class1, i, val, st. addclass(class1, fis0 + (i ' × val) ', st) = addclass(class1, fis0, st) ⊦ addclass(class1, ((sk × val0) ' + fis0) + (i ' × val) ', st) = addclass(class1, (sk × val0) ' + fis0, 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])]]