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Proof step 3 from proof lls-02 (previous step)

log2 m < log2 n, n ≤ m ⊦ n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
n ≠ 0 ⊦ ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1)
Proof step 5:
log2 m < log2 n, n ≤ m, ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1) ⊦ n = 0

Rule argument

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

Additional information

Used terms: n