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Proof step 20 from proof ac-dups-03 (previous step)

¬ (is_fskey((sk0 × val1) .1) ∧ (sk0 × val1) .1 .fs .class ≠ class1), is_fskey(sk0), ∀ class1, sk, val, fis1, val0, fis2, st. addclass(class1, fis0 + (sk × val) ' + fis1 + (sk × val0) ' + fis2, st) = addclass(class1, fis0 + (sk × val) ' + fis1 + fis2, st) ⊦ addclass(class1, ((sk0 × val1) ' + fis0) + (sk × val) ' + fis1 + (sk × val0) ' + fis2, st) = addclass(class1, ((sk0 × val1) ' + fis0) + (sk × val) ' + fis1 + fis2, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
is_fskey(sk0), ∀ class1, sk, val, fis1, val0, fis2, st. addclass(class1, fis0 + (sk × val) ' + fis1 + (sk × val0) ' + fis2, st) = addclass(class1, fis0 + (sk × val) ' + fis1 + fis2, st) ⊦ addclass(class1, ((sk0 × val1) ' + fis0) + (sk × val) ' + fis1 + (sk × val0) ' + fis2, st) = addclass(class1, ((sk0 × val1) ' + fis0) + (sk × val) ' + fis1 + fis2, st), ¬ is_indexkey(sk0) ∧ sk0 .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])]]