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Proof step 5 from proof lzer (previous step)

log2(m + 1) = 0, 2^ m0 < 2^(m0 + 1), m + 1 < 2^(m0 + 1), 2^ m0 ≤ m + 1 ⊦ m = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
⊦ 2^ m ≤ n ∧ n < 2^ m +1 → log2 n = m
Proof step 7:
log2(m + 1) = 0, m + 1 < 2^(m0 + 1), 2^ m0 < 2^(m0 + 1), 2^ m0 ≤ m + 1, 2^ m0 ≤ m + 1 ∧ m + 1 < 2^ m0 +1 → log2(m + 1) = m0 ⊦ m = 0

Rule argument

spec: <toplevel> name: Log2def seq: ⊦ 2^ m ≤ n ∧ n < 2^ m +1 → log2 n = m subst: [m, n] → [m0, m + 1]

Additional information

Used terms: m0, m + 1