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Proof step 2 from proof lookup-put-other (previous step)

st[r -- sks] .refval = st[r -- sks1] .refval, sks + sk ' ⊑ sks1, ∀ sks. sks + sk ' ⊑ sks1 → st[r -- sks] .refval ≠ st[r -- sks1] .refval ⊦ (st[st[r -- sks1] .refval -- sk, val][r -- (sks1 + sk ' + sks0)]) = st[r -- (sks1 + sk ' + sks0)]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 3, [QUANTTERMLIST : (sks)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks)]]