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Proof step 35 from proof exlist (previous step)

flatten(val ++ val1 ++ val2) = val3 ++ val0, ∀ val. ∃ val1, val3. flatten(val ++ val2) = val1 ++ val3, ∀ val. ∃ val3, val2. flatten(val ++ val1) = val3 ++ val2 ⊦ ∃ val0, val3. flatten(val ++ val1 ++ val2) = val0 ++ val3

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 1, [QUANTTERMLIST : (val3, val0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val3, val0)]]