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Proof step 38 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) ⊦ abs(j) = 1, j = 0, i0 >> (abs(j) - 1) = (2 * (i0 + 1) + j1 - 4 * 2 ^ (abs(j) - 2)) / (4 * 2 ^ (abs(j) - 2))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ j % 2 = 0 ∧ j ≠ 0 → i / j = (i / 2) / j / 2 subst: [j, i] → [4 * 2 ^ (abs(j) - 2), 2 * (i0 + 1) + j1 - 4 * 2 ^ (abs(j) - 2)] rotate: F spec: int-div inst: name: div-two-intro

Additional information

Used terms: 4 * 2 ^ (abs(j) - 2), 2 * (i0 + 1) + j1 - 4 * 2 ^ (abs(j) - 2)