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Proof step 54 from proof shiftright-neg-div (previous step)

i0 >> (abs(j) - 1) = (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), abs(j1) < 2, i0 ≤ 0, j1 ≤ 0, i0 * 2 + j1 < ~1, 1 < abs(j), 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < ~1, i0 * 2 + j1 < 0, 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < 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) ⊦ 2 ^ (abs(j) - 2) = 0, abs(j) = 1, j = 0, i0 >> (abs(j) - 1) = ((2 * (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) + j1) / 2) / (2 * 2 ^ (abs(j) - 2)), ¬ 2 * 2 ^ (abs(j) - 2) ≤ i0 + 1 ∧ j1 ≠ 0

User applied rule cut formula

And obtained the following new goals

Proof step 55:
i0 >> (abs(j) - 1) = (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), abs(j1) < 2, i0 ≤ 0, j1 ≤ 0, i0 * 2 + j1 < ~1, 1 < abs(j), 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < ~1, i0 * 2 + j1 < 0, 2 * (i0 + 2 ^ (abs(j) - 2)) + j1 < 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) ⊦ j1 = 0, 2 ^ (abs(j) - 2) = 0, abs(j) = 1, j = 0, i0 >> (abs(j) - 1) = ((2 * (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) + j1) / 2) / (2 * 2 ^ (abs(j) - 2)), ¬ 2 * 2 ^ (abs(j) - 2) ≤ i0 + 1 ∧ j1 ≠ 0

Rule argument

fma: j1 = 0

Additional information

Used simplifier rules: ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%j ≠ 0 ⊦ (i * j) / j = i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used formulas: j1 = 0