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Proof step 6 from proof int2bits-rec-pos (previous step)

0 ≤ i / 2, 0 ≤ i, 1 < i, 0 < i, 0 ≤ i ⊦ uint2bits(i) = uint2bits(i / 2) + (i % 2 = 1)'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
1 < i, 0 < i, 0 ≤ i ⊦ uint2bits(i) = uint2bits(i / 2) + (i % 2 = 1)'

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;