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Proof step 2 from proof numeral-06 (previous step)

m2 < m1 ⊦ m2 * m + n < m1 * m + n0 ↔ n < (m1 - m2) * m + n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m2 < m1 ⊦ m2 * m + n < m1 * m + n0 ↔ n < (m1 - m2) * m + n0, m1 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1