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Proof step 3 from proof keys-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ∃ ndx0, ndx1. [] = ndx0 + ndx1 ∧ ndx0.keys = [] ∧ ndx1.keys = []

Rule argument

Additional information

Used simplifier rules: keys-base-01: ⊦ [].keys = [];%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ kx + kx0 = [] ↔ kx = [] ∧ kx0 = []; (as)