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Proof step 9 from proof keys-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ndx0.keys = kx + kx1, ∀ 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 = nd.key ' + kx ∧ ndx1.keys = kx1

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ key0 ' + kx ↔ key = key0 ∨ key ∈ kx; (s)% ⊦ key ' + kx = key0 ' + kx0 ↔ key = key0 ∧ kx = kx0; (ws)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ key + kx = key ' + kx; (s)