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Proof step 19 from proof lsucc-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
4 * 2 ^ log2 n ≤ 2 * n + 1, 2 ^ log2 n * 2 ≤ n ⊦ n = 0, log2(2 * n + 1) = log2 n + 1

Rule argument

Additional information

Used simplifier rules: two-pot-zero: ⊦ 2 ^ n ≠ 0;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; from specification nat-mult%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult