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Proof step 4 from proof lookup-put-rkvs-notin (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ≠ rk0 → (st[rk0, val][rk]) = st[rk]])], [EXTRATERMS : (st[rk0, val], rk)]]