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Proof step 19 from proof ac-dups-03 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
sk0 = fs ', 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) ∧ fs .class ≠ class1

Rule argument

[speclemmaarg: ,is_fskey(sk) → (fs = sk .fs ↔ sk = fs ') ,[sk, fs] → [sk0, fs] ,T ,storekey , ,elim-03 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_fskey(sk) → (fs = sk .fs ↔ sk = fs ')])], [EXTRATERMS : (sk0, fs)]]