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Proof step 10 from proof ac-lookup-02 (previous step)

¬ (is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class2), 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) ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
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

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])]]