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Proof step 4 from proof keys-split (previous step)

⊦ ∃ ndx0, ndx1. [] = ndx0 + ndx1 ∧ ndx0.keys = [] ∧ ndx1.keys = []

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], []

Additional information

Used simplifier rules: keys-base-01: ⊦ [].keys = [];% ⊦ [] + ndx = ndx; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list] Used terms: [], []