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Proof step 19 from proof shiftright-neg-div (previous step)

0 ≤ 2 ^ (abs(j) - 2) + (i - 1) / 2, i + 2 ^ (abs(j) - 1) < 0, i < 0, abs(j) ≤ 1, ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0, (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

User applied rule cut formula

And closed the goal!

Rule argument

fma: abs(j) = 1

Additional information

Used simplifier rules: rightshift-zero: ⊦ i >> 0 = i;% ⊦ abs(i) ≤ 0 ↔ i = 0;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i = j → i - j = 0;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ (* i) + i0 - (* j) = (i - j) + i0;%i < 0 ⊦ (* i) + j = j - ~ i; Used formulas: abs(j) = 1