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Proof step 6 from proof nds-eq-one (previous step)

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

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

fma: nd0 ' + gndx0.nds = nd '

Additional information

Used simplifier rules: nds-eq-empty: ⊦ gndx.nds = [] ↔ gndx = [];%nds-base: ⊦ [].nds = [];% ⊦ gnd ' = gnd0 ' + gndx ↔ gnd = gnd0 ∧ gndx = []; (ws)%a: ⊦ (gndx + gndx0) + gndx1 = gndx + gndx0 + gndx1; from specification list[group-store]% ⊦ ndx + [] = ndx; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ mkgnode(nd, m, boolvar, boolvar0).nd = nd; (s)% ⊦ nd ' = nd0 ' ↔ nd = nd0; (s)% ⊦ nd ' = nd0 ' + ndx ↔ nd = nd0 ∧ ndx = []; (ws) Used formulas: nd0 ' + gndx0.nds = nd '