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Proof step 14 from proof ac-in-08 (previous step)

r ∈ st, ∀ r, st, class1. r ∈ st → r ∈ addclass(class1, fis0, st) ⊦ r ∈ addclass(class1, fv ' + fis0, st), fv .1 .fs .class ≠ class1 ∧ ¬ is_indexkey(fv .1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
fv = sk × val, r ∈ st, ∀ r, st, class1. r ∈ st → r ∈ addclass(class1, fis0, st) ⊦ r ∈ addclass(class1, fv ' + fis0, st), sk .fs .class ≠ class1 ∧ ¬ is_indexkey(sk)

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)]]