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Proof step 2 from proof addex (previous step)

m < n ⊦ n = 0, ∃ m0. m + m0 = n ∧ m0 ≠ 0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ m. m < n0 → n0 = 0 ∨ (∃ m0. m + m0 = n0 ∧ m0 ≠ 0), m < n0 +1 ⊦ n0 +1 = 0, ∃ m0. m + m0 = n0 +1 ∧ m0 ≠ 0

Rule argument

indvar: n

Additional information

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