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Proof step 5 from proof l-03 (previous step)

m < m0, ∀ n, m, m0. n < n1 ∧ m < m0 → m0 = 0 ∨ n1 = 0 ∨ m + n < m0 + n1, n = n1 ∨ n < n1 ⊦ m + n = n1 + m0, m0 = 0, m + n < n1 + m0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
m < m0, ∀ n, m, m0. n < n1 ∧ m < m0 → m0 = 0 ∨ n1 = 0 ∨ m + n < m0 + n1, n = n1 ∨ n < n1 ⊦ m + n = n1 + m0, m0 = 0, n1 = 0, m + n < n1 + m0

Rule argument

fma: n1 = 0

Additional information

Used simplifier rules: z-02: ⊦ 0 + m = m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%less-1: ⊦ ¬ n < 0; Used formulas: n1 = 0