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Proof step 2 from proof s-04 (previous step)

∀ m. m < n0 + 1 ↔ ¬ n0 < m ⊦ m < n0 +1 + 1 ↔ ¬ n0 +1 < m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ m. m < n0 + 1 ↔ ¬ n0 < m ⊦ m < 2 + n0 ↔ ¬ n0 + 1 < m

Rule argument

Additional information

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