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Proof step 38 from proof int2bytes-rec-pos (previous step)

8 = # fillsgn(9, uint2bits(i % 256)).rest, 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)) = [], bits2bytes(uint2bits(i / 256)) = bits2bytes(uint2bits(i / 256)) ∧ i2b(bits2int(fillsgn(9, uint2bits(i % 256)).rest)) = bits2bytes(int2bits(i % 256)).last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ # bits ≤ n → # fillsgn(n, bits) = n;% ⊦ bits ≠ [] → # bits.rest = (# bits)-1;