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Proof step 14 from proof ac-in-06 (previous step)

¬ (is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class2), is_fskey(sk), ∀ class1, class2, ty, fieldvar, st. class1 = class2 ∨ (jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st) ⊦ class1 = class2, jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, (sk × val) ' + fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
is_fskey(sk), ∀ class1, class2, ty, fieldvar, st. class1 = class2 ∨ (jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st) ⊦ class1 = class2, jvmref -- mkfs(class1, ty, fieldvar) ' ∈ addclass(class2, (sk × val) ' + fis0, st) ↔ jvmref -- mkfs(class1, ty, fieldvar) ' ∈ st, ¬ 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])]]