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Proof step 5 from proof s-04 (previous step)

∀ m. m < n0 + 1 ↔ ¬ n0 < m ⊦ m ≠ 2 + n0 ∧ ¬ 2 + n0 < m, n0 + 1 ≠ m ∧ ¬ m < n0 + 1

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

fma: m = 2 + n0

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%l: ⊦ m + n < m + n0 ↔ n < n0;%f: ⊦ m < n → n ≠ 0;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%oz: ⊦ m + 1 ≠ 0;%l: ⊦ m + n < m + n0 ↔ n < n0;%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1;%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1;%succ: ⊦ m +1 = m + 1; Used formulas: m = 2 + n0