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Proof step 4 from proof le (previous step)

m0 * m ≤ m0 * n, n < m ⊦ m0 = 0, m = 0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: numeral-03 seq: m1 < m2 ⊦ (* m2) * m + n ≤ (* m1) * m + n0 ↔ (m2 - m1) * m + n ≤ n0 subst: [m1, m2, m, n, n0] → [n, m, m0, 0, 0] flags: current, []

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m;%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%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%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%zero-01: ⊦ n < m → m - n ≠ 0; from specification nat Used terms: n, m, m0, 0, 0