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Proof step 2 from proof modp (previous step)

m < n, m0 * n ≤ m0 * m + n0, n0 < m0 ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ n, m0, n0. ¬(m1 < n ∧ m0 * n ≤ m0 * m1 + n0 ∧ n0 < m0), m1 +1 < n, m0 * n ≤ m0 * m1 +1 + n0, n0 < m0 ⊦

Rule argument

term: m

Additional information

Used simplifier rules: mult-base: ⊦ m * 0 = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%le-05: ⊦ m < n → (n * n0 ≤ m ↔ n0 = 0);%nat freely generated by 0, +1;;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2