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Proof step 31 from proof int2bits-add-low (previous step)

¬(j < 0 ∧ j0 < 0 ∧ abs(j0) < 2 ^ i ∧ 0 ≤ i), 0 < 2 ^ (i - 1) + j0, j0 < 0, j < 0, 0 < i, i < n→i(# int2bits(~(j * 2 ^ i + j0))), j * 2 ^ i + j0 < 0, 0 ≤ (j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0, j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i) ≤ 0 ⊦ j0 = 0, sgn(uint2bits((j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0)), j * 2 ^ i + j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
0 < 2 ^ (i - 1) + j0, j0 < 0, j < 0, 0 < i, i < n→i(# int2bits(~(j * 2 ^ i + j0))), j * 2 ^ i + j0 < 0, 0 ≤ (j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0, j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i) ≤ 0, 2 ^ i + j0 ≤ 0 ⊦ j0 = 0, sgn(uint2bits((j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0)), j * 2 ^ i + j0 = 0

Rule argument

Additional information

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