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Proof step 2 from proof numeral-15 (previous step)

m1 < m2 ⊦ m2 * m + n ≤ m1 * m ↔ (m2 - m1) * m + n ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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