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Proof step 10 from proof ac-bad-01 (previous step)

is_fskey(sk), ∀ class1, class2, ty, fieldvar, val, st. class1 = class2 ∨ addclass(class1, fis0 + (mkfs(class2, ty, fieldvar) ' × val) ', st) = addclass(class1, fis0, st) ⊦ class1 = class2, addclass(class1, ((sk × val0) ' + fis0) + (mkfs(class2, ty, fieldvar) ' × val) ', st) = addclass(class1, (sk × val0) ' + fis0, st), ¬ is_indexkey(sk) ∧ sk .fs .class ≠ class1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
sk = fs ', is_fskey(sk), ∀ class1, class2, ty, fieldvar, val, st. class1 = class2 ∨ addclass(class1, fis0 + (mkfs(class2, ty, fieldvar) ' × val) ', st) = addclass(class1, fis0, st) ⊦ class1 = class2, addclass(class1, ((sk × val0) ' + fis0) + (mkfs(class2, ty, fieldvar) ' × val) ', st) = addclass(class1, (sk × val0) ' + fis0, st), ¬ is_indexkey(sk) ∧ fs .class ≠ class1

Rule argument

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

Additional information

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