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Proof step 4 from proof keys-eq-one (previous step)

ndx0.keys = [] ⊦ ∃ nd. nd0 ' + ndx0 = nd ' ∧ nd.key = nd0.key

User applied rule constructor cut

And closed the goal!

Rule argument

var: ndx0

Additional information

Used simplifier rules: keys-base-01: ⊦ [].keys = [];%keys-rec-01: ⊦ (nd ' + ndx).keys = nd.key + ndx.keys;%node-list freely generated by [], +;;% ⊦ nd + ndx = nd ' + ndx; (s)% ⊦ key + kx = key ' + kx; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ kx + kx0 = [] ↔ kx = [] ∧ kx0 = []; (as)% ⊦ key ' ≠ []; (s)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ ndx + [] = ndx; (s)% ⊦ nd ' = nd0 ' ↔ nd = nd0; (s)