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Proof step 5 from proof lsucc-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
⊦ 2 ^ log2 n ≤ n, log2(2 * n + 1) = log2 n + 1, n = 0
Proof step 13:
⊦ 2 * n + 1 < 4 * 2 ^ log2 n, log2(2 * n + 1) = log2 n + 1, n = 0

Rule argument

pos: right 3

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