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Proof step 21 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(firstn(#(uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest) - 8, uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest)) + i2b(bits2int(lastn(8, 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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ bytes + by = bytes + by ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;