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Proof step 5 from proof int2bits-lslen-06 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 6:
m < # int2bits(i), 0 < i ⊦ ¬ # int2bits(i) ≤ i→n(n→i(m) - 1) + 1, m = 0
Proof step 9:
¬ # int2bits(i) ≤ i→n(n→i(m) - 1) + 1, 0 < i ⊦ m < # int2bits(i), m = 0

Rule argument

pos: right 1

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