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Proof step 14 from proof ac-refin (previous step)

rk ∈ st, ∀ rk, st, class. rk ∈ st → rk ∈ addclass(class, fis0, st) ⊦ rk ∈ addclass(class, fv ' + fis0, st), ¬ is_indexkey(fv .1) ∧ fv .1 .fs .class ≠ class

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
fv = sk × val, rk ∈ st, ∀ rk, st, class. rk ∈ st → rk ∈ addclass(class, fis0, st) ⊦ rk ∈ addclass(class, fv ' + fis0, st), ¬ is_indexkey(sk) ∧ sk .fs .class ≠ class

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)]]