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Proof step 4 from proof lls-01 (previous step)

2 * 2 ^ log2 n ≤ n, ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
log2 n < m, 2 ^ m < 2 * 2 ^ m, n < 2 * 2 ^ m, 2 * 2 ^ log2 n ≤ n, 2 ^ m ≤ n ⊦

Rule argument

Additional information

Used simplifier rules: pot2-def: ⊦ 2^ m = 2 ^ m;%lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%two-pot-zero: ⊦ 2 ^ n ≠ 0;%pot-mon: 1 < m ⊦ m ^ n < m ^ n0 ↔ n < n0;% ⊦ m < n → n ≠ 0; (s)% ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ 2 * m = 0 ↔ m = 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic