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Proof step 6 from proof lls (previous step)

m < log2 n, n < 2 ^ log2 n, n < 2 ^ (m + 1), 2 ^ m ≤ n ⊦ n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
⊦ 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;% ⊦ m +1 = m + 1; (s)% ⊦ ¬ n < n; (s) Used terms: m, n