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Proof step 3 from proof keys-eq-one (previous step)

ndx.keys = key ', ndx = nd0 + ndx0 ⊦ ∃ nd. ndx = nd ' ∧ nd.key = key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ndx0.keys = [] ⊦ ∃ nd. nd0 ' + ndx0 = nd ' ∧ nd.key = nd0.key

Rule argument

Additional information

Used simplifier rules: keys-rec-01: ⊦ (nd ' + ndx).keys = nd.key + ndx.keys;%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ key ' = key0 ' + kx ↔ key = key0 ∧ kx = []; (ws)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ key + kx = key ' + kx; (s)% ⊦ nd + ndx = nd ' + ndx; (s)