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Proof step 2 from proof modls (previous step)

n0 < m0 ⊦ m0 * m + n0 < m0 * n ↔ m < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < m0 ⊦ m0 * m + n0 < m0 * n ↔ m < n

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic