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Proof step 60 from proof shiftright-neg-div (previous step)

i0 >> (abs(j) - 1) = (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), 2 * 2 ^ (abs(j) - 2) ≤ i0 + 1, 1 < abs(j), i0 + 2 ^ (abs(j) - 2) < 0, i0 < 0, ∀ 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, abs(j) = 1, 2 ^ (abs(j) - 2) = 0, i0 >> (abs(j) - 1) = ((2 * (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) - 1) / 2) / (2 * 2 ^ (abs(j) - 2)), i0 ≠ ~1 ∧ ¬ abs(j) < 2

User applied rule cut formula

And closed the goal!

Rule argument

fma: i0 = ~1

Additional information

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