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Proof step 16 from proof int2bytes-rec-pos (previous step)

¬(uint2bits(i % 256) ≠ [] ∧ # uint2bits(i % 256) < 9), uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, 255 < i, 0 < i, 0 ≤ i, # uint2bits(i / 256) + # fillsgn(9, uint2bits(i % 256)).rest ≤ 8 ⊦ 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 17:
uint2bits(i) = uint2bits(i / 256) + uint2bits(i % 256).rest, 9 ≤ # uint2bits(i % 256), 255 < i, 0 < i, 0 ≤ i, # uint2bits(i % 256).rest + # uint2bits(i / 256) ≤ 8 ⊦ bits2bytes(uint2bits(i)) = bits2bytes(uint2bits(i / 256)) + bits2bytes(int2bits(i % 256)).last '

Rule argument

Additional information

Used simplifier rules: ⊦ n ≤ # bits → fillsgn(n, bits) = bits;% ⊦ ¬ m < n ↔ n ≤ m;%0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;