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Proof step 5 from proof addobj-at-other-03 (previous step)

∀ sk, r, class, st, r0. sk = _type ∨ (addobj(r, class, fis0, st)[r0 -- sk]) = st[r0 -- sk] ∨ sk ∈ fis0 ⊦ sk = _type, (addobj(r, class, fis0, st)[r -- fv .1, fv .2][r0 -- sk]) = st[r0 -- sk], sk ∈ fv ', sk ∈ fis0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
fv = sk0 × val, ∀ sk, r, class, st, r0. sk = _type ∨ (addobj(r, class, fis0, st)[r0 -- sk]) = st[r0 -- sk] ∨ sk ∈ fis0 ⊦ sk = _type, (addobj(r, class, fis0, st)[r -- sk0, val][r0 -- sk]) = st[r0 -- sk], sk ∈ fv ', sk ∈ fis0

Rule argument

[speclemmaarg: ,sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val ,[fv, sk, val] → [fv, sk0, val] ,T ,pair ,fieldinit ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val])], [EXTRATERMS : (fv, sk0, val)]]