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Proof step 5 from proof at-put-other-02 (previous step)

∀ st, r, sk, val, r0. (st[r -- sk, val][r0 -- sks0]) = st[r0 -- sks0] ∨ sk ∈ sks0 ⊦ sks0 = [], (st[r -- sk, val][st[r0 -- sk0] .refval -- sks0]) = st[st[r0 -- sk0] .refval -- sks0], sk = sk0, sk ∈ sks0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [QUANTTERMLIST : (st, r, sk, val, st[r0 -- sk0] .refval)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r, sk, val, st[r0 -- sk0] .refval)]]