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Proof step 15 from proof le-div-to-mult (previous step)

m = m0 +1, (m1 + m) * (n0 + m2) ≤ m1 * (n0 + m2) + n0 ⊦ m = 0, m2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
(n0 + m2) * (m1 + m0) + m2 ≤ m1 * (n0 + m2) ⊦ m2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m * (n + 1) = m * n + m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ m +1 = m + 1; (s)