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Proof step 12 from proof bint-and-011x-bits2unit (previous step)

abs(i + 1) < 4294967296, ~2147483648 ≤ i, i < 0 ⊦ bits2uint(fillsgn(32, int2bits(i))) = 4294967295 + i + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
~2147483648 ≤ i, i < 0 ⊦ bits2uint(fillsgn(32, int2bits(i))) = 4294967296 + i

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~ i - j = ~(i + j);% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i < 0 → abs(i + 1) = abs(i) - 1;