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Proof step 8 from proof rounded-help2 (previous step)

n0 < m1 + m, 2 * (m1 + m) ≤ n0 + m1 ⊦ m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [n, m, m0] → [m, m1, 2] flags: <all paths>

Additional information

Used simplifier rules: m2 ≠ 0 ⊦ (* m2) * m + n ≤ m + n0 ↔ (m2 - 1) * m + n ≤ n0; (s)% ⊦ m0 * 1 = m0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; (ws) Used terms: m, m1, 2