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Proof step 2 from proof less (previous step)

n < m ⊦ n + n0 < m + m0, m0 < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < m ⊦ m = 0, m0 ≠ n0 ∧ ¬ n0 < m0, n + n0 ≠ m + m0 ∧ ¬ m + m0 < n + n0

Rule argument

Additional information

Used simplifier rules: ls-04: ⊦ m < n ↔ ¬(m = n ∨ n < m);%f: ⊦ m < n → n ≠ 0;