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

n < 2 * 2^ m, 2^ m ≤ n, 2 * 2^ log2 n ≤ n, 2^ m < 2 * 2^ m, log2 n < m ⊦ 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] → [m, n] flags: current, []

Additional information

Used simplifier rules: lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%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%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: m, n