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Proof step 55 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, # uint2bits(i % 256) ≤ 9, 255 < i, 0 < i, 0 ≤ i, i % 256 ≤ 127 ⊦ uint2bits(i % 256).first ' + fillsgn(9 -1, uint2bits(i % 256)) = [], i2b(bits2int((uint2bits(i % 256).first ' + fillsgn(9 -1, uint2bits(i % 256))).rest)) = i2b(i % 256), bits2int((uint2bits(i % 256).first ' + fillsgn(9 -1, uint2bits(i % 256))).rest) = i % 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
uint2bits(i) = uint2bits(i / 256) + fillsgn(8, uint2bits(i % 256)), i % 256 ≤ 127, 0 ≤ i, 0 < i, 255 < i, # uint2bits(i % 256) < 9 ⊦ bits2int(uint2bits(i % 256)) = i % 256, i2b(bits2int(uint2bits(i % 256))) = i2b(i % 256)

Rule argument

Additional information

Used simplifier rules: ⊦ i2b(i) ≠ i2b(j) → i ≠ j; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n → m ≤ n;%0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;% ⊦ bits ≠ [] → bits2int(fillsgn(n, bits)) = bits2int(bits);% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;