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Proof step 22 from proof intshiftrsign-abs-le (previous step)

2 ^ band(j, 31) % 2 ^ band(j, 31) = 0 ∧ 0 ≤ 2 ^ band(j, 31) ∧ 0 ≤ ~(i + 1) ∧ 0 < 2 ^ band(j, 31), i + 1 < 2 ^ band(j, 31), band(j, 31) < 32, i < 0, 0 < band(j, 31), 0 < (2 ^ band(j, 31) / 2 ^ band(j, 31) + ~(i + 1) / 2 ^ band(j, 31)) + i ⊦ band(j, 31) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: band-commutativity: ⊦ band(i, j) = band(j, i);% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ i + 0 = i;% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + ~ j = i - j;%i ≠ 0 ⊦ i / i = 1;%j ≠ 0 ⊦ ~ i / j = ~(i / j);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;%i ≠ 0 ⊦ i % i = 0;% ⊦ 0 < j → (i ^ j = 0 ↔ i = 0);