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Proof step 3 from proof modp (previous step)

n0 < m0, m1 +1 < n, m0 * n ≤ m0 * m1 +1 + n0, ∀ n, n0, m0. ¬(m1 < n ∧ n0 < m0 ∧ m0 * n ≤ m0 * m1 + n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < m0, m1 + 1 < n, m0 * n ≤ (m1 * m0 + m0) + n0, ∀ n1, n2, m. ¬ m1 < n1 ∨ ¬ n2 < m ∨ ¬ m * n1 ≤ m * m1 + n2 ⊦

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%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)