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Proof step 2 from proof lsucc-01 (previous step)

⊦ log2(2 * n + 1) = log2 n + 1, n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 2^ m ≤ n ∧ n < 2^ m +1 → log2 n = m
Proof step 4:
2^(log2 n + 1) ≤ 2 * n + 1 ∧ 2 * n + 1 < 2^ (log2 n + 1)+1 → log2(2 * n + 1) = log2 n + 1 ⊦ log2(2 * n + 1) = log2 n + 1, n = 0

Rule argument

spec: nat-pot inst: name: Log2def seq: ⊦ 2^ m ≤ n ∧ n < 2^ m +1 → log2 n = m subst: [m, n] → [log2 n + 1, 2 * n + 1] flags: current, []

Additional information

Used terms: log2 n + 1, 2 * n + 1