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Proof step 6 from proof lookup-put-rkvs (previous step)

∀ rk, rk0, st, val. rk = rk0 ∨ (st[rk, val][rkvs0][rk0]) = (st[rkvs0][rk0]) ⊦ rk1 = rk, (st[rk1, val0][rk, val][rkvs0][rk0]) = (st[rk1, val0][rkvs0][rk0]), rk = rk0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [QUANTTERMLIST : (rk, rk0, st[rk1, val0], val)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, rk0, st[rk1, val0], val)]]