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

log2 n < m, 2 ^ m < 2 * 2 ^ m, n < 2 * 2 ^ m, 2 * 2 ^ log2 n ≤ n, 2 ^ m ≤ n ⊦

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

Additional information

Used simplifier rules: pot2-def: ⊦ 2^ m = 2 ^ m;%lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;% ⊦ 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% ⊦ ¬ n < n; (s) Used terms: m, n