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Proof step 8 from proof keys-split (previous step)

∀ kx0, kx1. ndx0.keys = kx0 + kx1 → (∃ ndx2, ndx1. ndx0 = ndx2 + ndx1 ∧ ndx2.keys = kx0 ∧ ndx1.keys = kx1) ⊦ ∃ ndx, ndx1. nd ' + ndx0 = ndx + ndx1 ∧ ndx.keys = [] ∧ ndx1.keys = nd.key ' + ndx0.keys

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], nd ' + ndx0 abort: F discard: T computed: F

Additional information

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