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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < 2 * 2^ m, 2^ m ≤ n, 2 * 2^ log2 n ≤ n, 2^ m < 2 * 2^ m, log2 n < m ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%zero: ⊦ 2^ n ≠ 0;%potmon: ⊦ 2^ m < 2^ n ↔ m < n;%two-ls-01: ⊦ 2 * m < 2 * n ↔ m < n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%two-zero-01: ⊦ 2 * m = 0 ↔ m = 0; from specification nat-mult%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2