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Proof step 12 from proof keys-split (previous step)

⊦ ∃ ndx3, ndx2. nd ' + ndx + ndx1 = ndx3 + ndx2 ∧ ndx3.keys = nd.key ' + ndx.keys ∧ ndx2.keys = ndx1.keys

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: nd ' + ndx, ndx1

Additional information

Used simplifier rules: keys-rec-01: ⊦ (nd ' + ndx).keys = nd.key + ndx.keys;%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ key + kx = key ' + kx; (s)% ⊦ key ' + kx = key ' + kx0 ↔ kx = kx0; (s)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set] Used terms: nd ' + ndx, ndx1