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Proof step 25 from proof ac-mode-01 (previous step)

is_fskey(fv .1), ∀ class0, st0. (addclass(class0, fis0, st0)[_mode]) = st0[_mode] ⊦ fv .1 .fs .class = class, (addclass(class, fis0, st)[_mode]) = st[_mode], mkfs(`java.lang.Object' .class, void_type, `mode' .field) ' ∈ fis0, mkfs(`java.lang.Object' .class, void_type, `mode' .field) ' ∈ fv '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
fv = sk × val, is_fskey(sk), ∀ class0, st0. (addclass(class0, fis0, st0)[_mode]) = st0[_mode] ⊦ sk .fs .class = class, (addclass(class, fis0, st)[_mode]) = st[_mode], mkfs(`java.lang.Object' .class, void_type, `mode' .field) ' ∈ fis0, mkfs(`java.lang.Object' .class, void_type, `mode' .field) ' ∈ fv '

Rule argument

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

Additional information

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