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Proof step 4 from proof numeral-12 (previous step)

m1 = n1 + m2, m2 < m1, m2 ≤ m1 ⊦ m2 * m < m1 * m + n0 ↔ n1 * m + n0 ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m2 < n1 + m2 ⊦ m2 * m < (m * n1 + m * m2) + n0 ↔ n1 * m + n0 ≠ 0

Rule argument

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;% ⊦ m < n → m ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic