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Proof step 4 from proof nds-split (previous step)

gnd.nd ' + gndx0.nds = ndx0 + ndx1, ∀ ndx0, ndx1. gndx0.nds = ndx0 + ndx1 → (∃ gndx2, gndx1. gndx0 = gndx2 + gndx1 ∧ ndx0 = gndx2.nds ∧ ndx1 = gndx1.nds), gnd.nd ∈ ndx0 ∨ gnd.nd ∈ ndx1 ⊦ ∃ gndx, gndx1. gnd ' + gndx0 = gndx + gndx1 ∧ ndx0 = gndx.nds ∧ ndx1 = gndx1.nds

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
nd ' + gndx0.nds = ndx0 + ndx1, gnd = mkgnode(nd, m, boolvar, boolvar0), ∀ ndx0, ndx1. gndx0.nds = ndx0 + ndx1 → (∃ gndx2, gndx1. gndx0 = gndx2 + gndx1 ∧ ndx0 = gndx2.nds ∧ ndx1 = gndx1.nds), nd ∈ ndx0 ∨ nd ∈ ndx1 ⊦ ∃ gndx, gndx1. gnd ' + gndx0 = gndx + gndx1 ∧ ndx0 = gndx.nds ∧ ndx1 = gndx1.nds

Rule argument

spec: groupnode name: elim seq: ⊦ ((nd = gnd.nd ∧ m = gnd.sqnum) ∧ (boolvar ↔ gnd.start)) ∧ (boolvar0 ↔ gnd.end) ↔ gnd = mkgnode(nd, m, boolvar, boolvar0) subst: [gnd, nd, m, boolvar, boolvar0] → [gnd, nd, m, boolvar, boolvar0]

Additional information

Used simplifier rules: ⊦ ((nd = gnd.nd ∧ m = gnd.sqnum) ∧ (boolvar ↔ gnd.start)) ∧ (boolvar0 ↔ gnd.end) ↔ gnd = mkgnode(nd, m, boolvar, boolvar0); (elim) Used terms: gnd, nd, m, boolvar, boolvar0