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Proof step 4 from proof int2bits-leneq-pos (previous step)

0 < j ⊦ j = 0, 2 ^ (n→i(# int2bits(j)) - 2) ≤ j ∧ j < 2 ^ (n→i(# int2bits(j)) - 1)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
0 < j ⊦ 2 ^ (n→i(# int2bits(j)) - 2) ≤ j, j = 0
Proof step 13:
0 < j ⊦ j < 2 ^ (n→i(# int2bits(j)) - 1), j = 0

Rule argument

pos: right 2

Additional information