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Proof step 6 from proof lls-02 (previous step)

n < 2 ^ (m0 + 1), log2 m < log2 n, 2 ^ m0 ≤ m, 2 ^ m0 ≤ n, n ≤ m ⊦ n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
n ≠ 0 ⊦ ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1)
Proof step 8:
log2 m < log2 n, n < 2 ^ (m0 + 1), n ≤ m, 2 ^ m0 ≤ n, 2 ^ m0 ≤ m, ∃ m0. 2^ m0 ≤ m ∧ m < 2^(m0 + 1) ⊦ n = 0

Rule argument

spec: <toplevel> name: potex seq: n ≠ 0 ⊦ ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1) subst: [n] → [m]

Additional information

Used simplifier rules: ⊦ m ≤ 0 ↔ m = 0; (s) Used terms: m