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Proof step 6 from proof keys-split (previous step)

nd.key ' + ndx0.keys = kx0 + kx1, ∀ 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

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
nd.key ' + ndx0.keys = kx0 + kx1, kx0 = [], ∀ 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
Proof step 9:
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

Rule argument

var: kx0

Additional information

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