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Proof step 24 from proof ac-in-10 (previous step)

is_indexkey(fv .1), r -- sk ∈ addclass(class, fis0, st), r ∈ addclass(class, fis0, st), ∀ sk0, class0, r0, st0. ¬ sk0 ∈ fis0 ∧ sk0 ≠ mkfs(class0, void_type, `initstate' .field) ' → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ sk = mkfs(class, void_type, `initstate' .field) ', r -- sk ∈ st, sk ∈ fv ', sk ∈ fis0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 23:
fv = sk1 × val, is_indexkey(sk1), r -- sk ∈ addclass(class, fis0, st), r ∈ addclass(class, fis0, st), ∀ sk0, class0, r0, st0. ¬ sk0 ∈ fis0 ∧ sk0 ≠ mkfs(class0, void_type, `initstate' .field) ' → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ sk = mkfs(class, void_type, `initstate' .field) ', r -- sk ∈ st, sk ∈ fv ', sk ∈ fis0

Rule argument

[speclemmaarg: ,sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val ,[fv, sk, val] → [fv, sk1, val] ,T ,pair ,fieldinit ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val])], [EXTRATERMS : (fv, sk1, val)]]