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Proof step 10 from proof union-def (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 2, [QUANTTERMLIST : (rk, st, (st ∪ st1)[rk])]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, st, (st ∪ st1)[rk])]]