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Proof step 5 from proof nds-eq-one (previous step)

gnd = mkgnode(nd0, m, boolvar, boolvar0), ∀ nd. gndx0.nds = nd ' ↔ (∃ gnd. gndx0 = gnd ' ∧ gnd.nd = nd) ⊦ nd0 ' + gndx0.nds = nd ' ↔ (∃ gnd0. gnd ' + gndx0 = gnd0 ' ∧ gnd0.nd = nd)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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