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Proof step 4 from proof modls-03 (previous step)

n1 < n, m0 +1 < m, n2 < n, m * n + n2 ≤ m0 +1 * n + n1, ∀ n2, n, m, n1. ¬(n2 < n ∧ m0 < m ∧ n1 < n ∧ m * n + n2 ≤ m0 * n + n1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n1 < n, n2 < n, m0 + 1 < m, m * n + n2 ≤ (n * m0 + n) + n1, ∀ n0, n3, m1, n4. ¬ n0 < n3 ∨ ¬ m0 < m1 ∨ ¬ n4 < n3 ∨ ¬ m1 * n3 + n0 ≤ m0 * n3 + n4 ⊦

Rule argument

Additional information

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