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Proof step 4 from proof l (previous step)

∀ n, n0. (∀ m1, n, n0. m1 < m0 → true → (m1 + n < m1 + n0 ↔ n < n0)) → (m0 + n < m0 + n0 ↔ n < n0), ∀ m1, n, n0. m1 < m0 +1 → true → (m1 + n < m1 + n0 ↔ n < n0) ⊦ m0 +1 + n < m0 +1 + n0 ↔ n < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ m, n1, n2. ¬ m0 < m → (m + n1 < m + n2 ↔ n1 < n2), (m < m0 → (m + n1 < m + n2 ↔ n1 < n2)) → (∀ n1, n2. m0 + n1 < m0 + n2 ↔ n1 < n2) ⊦ m0 + n < m0 + n0 ↔ n < n0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%s: ⊦ m < n + 1 ↔ ¬ n < m;%c: ⊦ m + n = n + m;