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Proof step 4 from proof s-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ m. m < n0 + 1 ↔ ¬ n0 < m ⊦ m ≠ 2 + n0 ∧ ¬ 2 + n0 < m, n0 + 1 ≠ m ∧ ¬ m < n0 + 1

Rule argument

Additional information

Used simplifier rules: ls-04: ⊦ m < n ↔ ¬(m = n ∨ n < m);