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Proof step 2 from proof q (previous step)

⊦ n = 1, n = 0, ∃ m. m +1 +1 = n

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
n0 = 1 ∨ n0 = 0 ∨ (∃ m. m +1 +1 = n0) ⊦ n0 +1 = 1, n0 +1 = 0, ∃ m. m +1 +1 = n0 +1

Rule argument

term: n

Additional information

Used simplifier rules: gen: ⊦ ;