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Proof step 9 from proof uint2bits-add-high-01 (previous step)

j0 < 0 ∧ 0 < i ∧ 2 ^ (i - 1) < abs(j0) ∧ abs(j0) ≤ 2 ^ i, 1 < j, 0 < j, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 ≤ j0 + 2 ^ i, 0 ≤ i ⊦ fillsgn(i→n(i) + 1, sgn(int2bits(j0).rest)).rest = fillsgn(i→n(i), int2bits(j0).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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