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Proof step 20 from proof ac-in-11 (previous step)

r -- sk ∈ addclass(class, fv ' + fis0, st), is_indexkey(sk), r ∈ addclass(class, fv ' + fis0, st), is_fskey(fv .1), ∀ sk0, r0, class0, st0. ¬ is_fskey(sk0) → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ r -- sk ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
fv = sk1 × val, r -- sk ∈ addclass(class, fv ' + fis0, st), is_indexkey(sk), r ∈ addclass(class, fv ' + fis0, st), is_fskey(sk1), ∀ sk0, r0, class0, st0. ¬ is_fskey(sk0) → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ r -- sk ∈ st

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)]]