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Proof step 4 from proof addobj-at-other-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ sk, r, class, st, r0. sk = _type ∨ (addobj(r, class, fis0, st)[r0 -- sk]) = st[r0 -- sk] ∨ sk ∈ fis0 ⊦ sk = sk0, sk = _type, (addobj(r, class, fis0, st)[r0 -- sk]) = st[r0 -- sk], sk ∈ fis0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ≠ sk0 → (st[r -- sk, val][r0 -- sk0]) = st[r0 -- sk0]], [CSIMPSEQ : ⊦ sk ∈ (sk0 × val) ' ↔ sk = sk0])]]