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Proof step 6 from proof int2bits-len-pos-pred (previous step)

0 ≤ i, 0 < j, j < 2 ^ i * 2, 2 ^ i ≤ j ⊦ 0 = j, # int2bits(j - 1) = # int2bits(j), 2 ^ i = j, ∃ i. 2 ^ i = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ i, 0 < j, j < 2 ^ i * 2, 2 ^ i < j ⊦ 0 = j, # int2bits(j - 1) = # int2bits(j), 2 ^ i = j, ∃ i. 2 ^ i = j

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);