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

Additional information

Used simplifier rules: pot2-def: ⊦ 2^ m = 2 ^ m;%lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%pot-mon: 1 < m ⊦ m ^ n < m ^ n0 ↔ n < n0;%two-pot-zero: ⊦ 2 ^ n ≠ 0;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ 2 * n ≤ n ↔ n = 0; (s)%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