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Proof step 2 from proof nds-eq-one (previous step)

⊦ gndx.nds = nd ' ↔ (∃ gnd. gndx = gnd ' ∧ gnd.nd = nd)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ nd. gndx0.nds = nd ' ↔ (∃ gnd. gndx0 = gnd ' ∧ gnd.nd = nd) ⊦ (gnd + gndx0).nds = nd ' ↔ (∃ gnd0. gnd + gndx0 = gnd0 ' ∧ gnd0.nd = nd)

Rule argument

term: gndx

Additional information

Used simplifier rules: nds-base: ⊦ [].nds = [];%group-node-list freely generated by [], +;;% ⊦ nd ' ≠ []; (s)% ⊦ gnd ' ≠ []; (s)