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Proof step 9 from proof ac-lookup-02 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
sk = fs ', is_fskey(sk), ∀ class1, class2, st, ty, fieldvar. class1 = class2 ∨ (addclass(class2, fis0, st)[jvmref -- mkfs(class1, ty, fieldvar) ']) = st[jvmref -- mkfs(class1, ty, fieldvar) '] ⊦ class1 = class2, (addclass(class2, (sk × val) ' + fis0, st)[jvmref -- mkfs(class1, ty, fieldvar) ']) = st[jvmref -- mkfs(class1, ty, fieldvar) '], ¬ is_indexkey(sk) ∧ fs .class ≠ class2

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)]]