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Proof step 3 from proof union-at-left (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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

Additional information

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