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Proof step 13 from proof int2bytes-rec-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
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 '

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;