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Proof step 10 from proof okfieldinits-index (previous step)

is_indexkey(sk), okfieldinits(class, fv ' + fis0), ∀ sk0, class0. ¬ sk0 ∈ fis0 .keys ∨ ¬ okfieldinits(class0, fis0) ∨ is_fskey(sk0) ⊦ sk ≠ fv .1 ∧ ¬ sk ∈ fis0 .keys

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
fv = sk1 × val, is_indexkey(sk), okfieldinits(class, fv ' + fis0), ∀ sk0, class0. ¬ sk0 ∈ fis0 .keys ∨ ¬ okfieldinits(class0, fis0) ∨ is_fskey(sk0) ⊦ sk ≠ sk1 ∧ ¬ sk ∈ fis0 .keys

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)]]