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Proof step 10 from proof keys-split (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
ndx0.keys = kx + kx1, ndx0.keys = kx + kx1 → (∃ ndx2, ndx1. ndx0 = ndx2 + ndx1 ∧ ndx2.keys = kx ∧ ndx1.keys = kx1) ⊦ ∃ ndx, ndx1. nd ' + ndx0 = ndx + ndx1 ∧ ndx.keys = nd.key ' + kx ∧ ndx1.keys = kx1

Rule argument

pos: left 2 terms: kx, kx1 abort: F discard: T computed: T

Additional information

Used terms: kx, kx1