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Proof step 2 from proof lls (previous step)

n < 2^ log2 n ⊦ n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
n ≠ 0 ⊦ ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1)
Proof step 4:
∃ m. 2^ m ≤ n ∧ n < 2^(m + 1), n < 2^ log2 n ⊦ n = 0

Rule argument

spec: nat-pot inst: name: potex seq: n ≠ 0 ⊦ ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1) subst: [n] → [n] flags: current, []

Additional information

Used terms: n