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Proof step 18 from proof int2bytes-rec-pos (previous step)

uint2bits(i % 256) ≠ [] ∧ # uint2bits(i % 256) < 9, uint2bits(i) = uint2bits(i / 256) + (uint2bits(i % 256).first ' + fillsgn(9 -1, uint2bits(i % 256))).rest, 255 < i, 0 < i, 0 ≤ i, # uint2bits(i / 256) + # (uint2bits(i % 256).first ' + fillsgn(9 -1, 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 19:
uint2bits(i) = uint2bits(i / 256) + fillsgn(8, uint2bits(i % 256)), # fillsgn(8, uint2bits(i % 256)) + # uint2bits(i / 256) ≤ 8, 0 ≤ i, 0 < i, 255 < i, # uint2bits(i % 256) < 9 ⊦ bits2bytes(uint2bits(i)) = bits2bytes(uint2bits(i / 256)) + bits2bytes(int2bits(i % 256)).last '

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)%0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;