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Proof step 5 from proof lzer (previous step)

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

Rule argument

spec: nat-pot inst: name: Log2def seq: ⊦ 2^ m ≤ n ∧ n < 2^ m +1 → log2 n = m subst: [m, n] → [m0, m + 1] flags: current, []

Additional information

Used simplifier rules: lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%Pot2-zero: ⊦ 2^ 0 = 1;%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%lep: ⊦ m ≤ m + n; from specification nat%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%l-num-05: n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1 Used terms: m0, m + 1