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Proof step 29 from proof exlist (previous step)

⊦ ∃ val0, val1. flatten(val) ++ doubleval(doublevar) = val0 ++ val1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 1, [QUANTTERMLIST : (flatten(val), doubleval(doublevar))]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (flatten(val), doubleval(doublevar))]]