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Proof step 7 from proof put-put-switch (previous step)

∀ st, r, sk, val, r0, val0. (st[r -- sk, val][r0 -- sks, val0]) = (st[r0 -- sks, val0][r -- sk, val]) ∨ sk ∈ sks ⊦ sks = [], (st[r -- sk, val][st[r0 -- sk0] .refval -- sks, val0]) = (st[st[r0 -- sk0] .refval -- sks, val0][r -- sk, val]), sk = sk0, sk ∈ sks

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, val0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r, sk, val, st[r0 -- sk0] .refval, val0)]]