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

⊦ ∃ gndx3, gndx2. mkgnode(nd0, m, boolvar, boolvar0) ' + gndx + gndx1 = gndx3 + gndx2 ∧ nd0 ' + gndx.nds = gndx3.nds ∧ gndx1.nds = gndx2.nds

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: mkgnode(nd0, m, boolvar, boolvar0) ' + gndx, gndx1

Additional information

Used simplifier rules: a: ⊦ (gndx + gndx0) + gndx1 = gndx + gndx0 + gndx1; from specification list[group-store]% ⊦ (gnd ' + gndx).nds = gnd.nd + gndx.nds; (s)% ⊦ mkgnode(nd, m, boolvar, boolvar0).nd = nd; (s)% ⊦ nd + ndx = nd ' + ndx; (s)% ⊦ nd ' + ndx = nd ' + ndx0 ↔ ndx = ndx0; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list] Used terms: mkgnode(nd0, m, boolvar, boolvar0) ' + gndx, gndx1