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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [QUANTTERMLIST : (sk, r, class, st, r0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, r, class, st, r0)]]