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Proof step 4 from proof addex (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 5:
∀ m. m < n0 → n0 = 0 ∨ (∃ m0. m + m0 = n0 ∧ m0 ≠ 0) ⊦ n0 = 0, n0 < m, ∃ m0. m + m0 = n0 + 1 ∧ m0 ≠ 0

Rule argument

fma: n0 = 0

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%one-01: ⊦ n ≠ 0 → (m + n = 1 ↔ n = 1 ∧ m = 0);%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%zero-01: ⊦ ¬ m < 0; from specification nat-basic1 Used formulas: n0 = 0