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Proof step 4 from proof ltwo-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ log2(2 * n) = log2 n + 1, n = 0, 2 * 2 ^ log2 n ≤ 2 * n ∧ 2 * n < 4 * 2 ^ log2 n

Rule argument

Additional information

Used simplifier rules: lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%pot2-def: ⊦ 2^ m = 2 ^ m;%add: ⊦ m ^ (n + n0) = m ^ n * m ^ n0;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2