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Proof step 7 from proof le (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: numeral-05 seq: m1 < m2 ⊦ (* m2) * m + n < (* m1) * m + n0 ↔ (m2 - m1) * m + n < n0 subst: [m1, m2, m, n, n0] → [m, n, m0, 0, 0] flags: current, []

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m;%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%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%ler: ⊦ m ≤ m; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: m, n, m0, 0, 0