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Proof step 21 from proof int2bits-rec-neg (previous step)

0 ≤ ~ i, i < 0 ⊦ sgn(bitsinvert(sgn(bitsinvert(bitsinvert(uint2bits(~ i + 1)) +1))) + true ') = sgn(bitsinvert(uint2bits(1 - i)) +1) + true ', i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
i < 0 ⊦ i = 0, sgn(bitsinvert(sgn(bitsinvert(bitsinvert(uint2bits(1 - i)) +1))) + true ') = sgn(bitsinvert(uint2bits(1 - i)) +1) + true '

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i + 0 = i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;