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Proof step 3 from proof nds-eq-one (previous step)

∀ nd. gndx0.nds = nd ' ↔ (∃ gnd. gndx0 = gnd ' ∧ gnd.nd = nd) ⊦ (gnd + gndx0).nds = nd ' ↔ (∃ gnd0. gnd + gndx0 = gnd0 ' ∧ gnd0.nd = nd)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ nd. gndx0.nds = nd ' ↔ (∃ gnd. gndx0 = gnd ' ∧ gnd.nd = nd) ⊦ gnd.nd ' + gndx0.nds = nd ' ↔ (∃ gnd0. gnd ' + gndx0 = gnd0 ' ∧ gnd0.nd = nd)

Rule argument

Additional information

Used simplifier rules: nds-rec: ⊦ (gnd ' + gndx).nds = gnd.nd + gndx.nds;%a: ⊦ (gndx + gndx0) + gndx1 = gndx + gndx0 + gndx1; from specification list[group-store]% ⊦ nd + ndx = nd ' + ndx; (s)% ⊦ gnd + gndx = gnd ' + gndx; (s)