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Proof step 2 from proof keys-split (previous step)

ndx.keys = kx0 + kx1 ⊦ ∃ ndx0, ndx1. ndx = ndx0 + ndx1 ∧ ndx0.keys = kx0 ∧ ndx1.keys = kx1

User applied rule structural induction

And obtained the following new goals

Proof step 3:
[].keys = kx0 + kx1 ⊦ ∃ ndx0, ndx1. [] = ndx0 + ndx1 ∧ ndx0.keys = kx0 ∧ ndx1.keys = kx1
Proof step 5:
(nd + ndx0).keys = kx0 + kx1, ∀ kx0, kx1. ndx0.keys = kx0 + kx1 → (∃ ndx2, ndx1. ndx0 = ndx2 + ndx1 ∧ ndx2.keys = kx0 ∧ ndx1.keys = kx1) ⊦ ∃ ndx2, ndx1. nd + ndx0 = ndx2 + ndx1 ∧ ndx2.keys = kx0 ∧ ndx1.keys = kx1

Rule argument

indvar: ndx

Additional information

Used simplifier rules: node-list freely generated by [], +;;