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Proof step 2 from proof int2bits-lslen-12 (previous step)

0 < i, 0 ≤ j ⊦ 2 ^ j < i ↔ ¬ # int2bits(i) < i→n(j) + 2 ∧ i ≠ 2 ^ j

User applied rule case distinction

And obtained the following new goals

Proof step 3:
2 ^ j < i, 0 < i, 0 ≤ j ⊦ ¬ # int2bits(i) < i→n(j) + 2 ∧ i ≠ 2 ^ j
Proof step 6:
¬ # int2bits(i) < i→n(j) + 2 ∧ i ≠ 2 ^ j, 0 < i, 0 ≤ j ⊦ 2 ^ j < i

Rule argument

pos: right 1

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