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Proof step 3 from proof int2bits-lelen (previous step)

n→i(m) = i0 + 1, 0 < i ⊦ # int2bits(i) ≤ m ↔ ¬ 2 ^ i0 ≤ i

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < j → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1))
Proof step 5:
0 < i → (∃ i0. 0 ≤ i0 ∧ 2 ^ i0 ≤ i ∧ i < 2 ^ (i0 + 1)), n→i(m) = i0 + 1, 0 < i ⊦ # int2bits(i) ≤ m ↔ ¬ 2 ^ i0 ≤ i

Rule argument

seq: ⊦ 0 < j → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)) subst: [j] → [i] rotate: F spec: int-pot inst: name: exin2pot

Additional information

Used terms: i