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Proof step 2 from proof validrefkeynotnull-union-01 (previous step)

validrefkeynotnull(str, rk, ty, st) ⊦ validrefkeynotnull(str, rk, ty, st0 ∪ st), ∃ st2. st0 ∪ st = st2 ∪ st

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 2, [QUANTTERMLIST : (st0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0)]]