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Proof step 43 from proof int2bytes-rec-pos (previous step)

¬(127 < i % 256 ∧ i % 256 ≤ 255), uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, # uint2bits(i % 256) ≤ 9, 255 < i, 0 < i, 0 ≤ i ⊦ fillsgn(9, uint2bits(i % 256)) = [], i2b(bits2int(fillsgn(9, uint2bits(i % 256)).rest)) = int2bytes(i % 256).last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, # uint2bits(i % 256) ≤ 9, 255 < i, 0 < i, 0 ≤ i, i % 256 ≤ 127 ⊦ fillsgn(9, uint2bits(i % 256)) = [], i2b(bits2int(fillsgn(9, uint2bits(i % 256)).rest)) = int2bytes(i % 256).last

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ 255 < i % 256;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;