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Proof step 1 from proof nds-split (conclusion)

⊦ gndx.nds = ndx0 + ndx1 ↔ (∃ gndx0, gndx1. gndx = gndx0 + gndx1 ∧ ndx0 = gndx0.nds ∧ ndx1 = gndx1.nds)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
gndx.nds = ndx0 + ndx1 ⊦ ∃ gndx0, gndx1. gndx = gndx0 + gndx1 ∧ ndx0 = gndx0.nds ∧ ndx1 = gndx1.nds

Rule argument

Additional information

Used simplifier rules: a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ ndx + ndx0 = ndx + ndx1 ↔ ndx0 = ndx1; (s)%a: ⊦ (gndx + gndx0) + gndx1 = gndx + gndx0 + gndx1; from specification list[group-store]% ⊦ (gndx0 + gndx1).nds = gndx0.nds + gndx1.nds; (s)