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Proof step 18 from proof ltwo-03 (previous step)

n < 2 * 2 ^ log2 n, 4 * 2 ^ log2 n ≤ 2 * n ⊦ log2(2 * n) = log2 n + 1, n = 0

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

spec: nat-mult name: le seq: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n) subst: [m0, m, n] → [2, 2 * 2 ^ log2 n, n]

Additional information

Used simplifier rules: pot-mon: 1 < m ⊦ m ^ n < m ^ n0 ↔ n < n0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws)% ⊦ ¬ n < n; (s) Used terms: 2, 2 * 2 ^ log2 n, n