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Proof step 4 from proof union-at-right-01 (previous step)

rk ∈ st1, ∀ rk, st. rk ∈ st1 → (st ∪ st1)[rk] = st1[rk] ⊦ rk = rk0, (st ∪ st1)[rk] = st1[rk]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 2, [QUANTTERMLIST : (rk, st)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, st)]]