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Proof step 9 from proof int2bytes-rec-pos (previous step)

0 ≤ i / 256, uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, 255 < i, 0 < i, 0 ≤ i ⊦ bits2bytes(uint2bits(i)) = bits2bytes(uint2bits(i / 256)) + bits2bytes(int2bits(i % 256)).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, 255 < i, 0 < i, 0 ≤ i ⊦ bits2bytes(uint2bits(i)) = bits2bytes(uint2bits(i / 256)) + bits2bytes(int2bits(i % 256)).last '

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i / j;