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Proof step 2 from proof lpot (previous step)

⊦ log2 2^ n = n

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 2^ m ≤ n ∧ n < 2^ m +1 → log2 n = m

Rule argument

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

Additional information

Used simplifier rules: potmonleq: ⊦ 2^ m ≤ 2^ n ↔ m ≤ n;%lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%zero: ⊦ 2^ n ≠ 0;%ler: ⊦ m ≤ m; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%twole: ⊦ 2 * n ≤ n ↔ n = 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n, 2^ n