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Proof step 2 from proof three (previous step)

m + n = 3 ⊦ m = 1, n = 0, m = 0, m = 2 ∧ n = 1

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ m. m + n0 = 3 → m = 1 ∨ n0 = 0 ∨ m = 0 ∨ m = 2 ∧ n0 = 1, m + n0 +1 = 3 ⊦ m = 1, n0 +1 = 0, m = 0, m = 2 ∧ n0 +1 = 1

Rule argument

term: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;