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Proof step 80 from proof shiftright-rec-8-neg (previous step)

j0 < 0, j0 * 256 >> (j - 1) = (j0 * 256 - 255) / 256 >> (j - 9), (i - 1) / 2 = j0 * 256, 8 < j, i < 0, 0 < j, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 8, i % 256 = 0, (j0 * 256 - 255) / 256 = (i / 256 - 2) / 2, j = 0, j0 * 256 >> (j - 1) = (i / 256 - 2) / 2 >> (j - 9)

User applied rule cut formula

And obtained the following new goals

Proof step 81:
i / 256 - 2 = ~(2 + ~(i / 256)), j0 < 0, j0 * 256 >> (j - 1) = (j0 * 256 - 255) / 256 >> (j - 9), (i - 1) / 2 = j0 * 256, 8 < j, i < 0, 0 < j, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 8, i % 256 = 0, (j0 * 256 - 255) / 256 = (i / 256 - 2) / 2, j = 0, j0 * 256 >> (j - 1) = (i / 256 - 2) / 2 >> (j - 9)

Rule argument

fma: i / 256 - 2 = ~(2 + ~(i / 256))

Additional information

Used simplifier rules: ⊦ i + ~ j = i - j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ~(i - j) = j - i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + (j - i) = j; Used formulas: i / 256 - 2 = ~(2 + ~(i / 256))