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Proof step 36 from proof shiftright-neg-div (previous step)

(i - 1) / 2 >> (abs(j) - 1) = ((i - 1) / 2 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), 2 * 2 ^ (abs(j) - 2) + i < 0, 1 < abs(j), i < 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) ⊦ (i - 1) / 2 >> (abs(j) - 1) = (i + 1 - 4 * 2 ^ (abs(j) - 2)) / (4 * 2 ^ (abs(j) - 2)), j = 0, abs(j) = 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 37:
i - 1 = i0 * 2 + j1 ∧ j1 ≤ 0 ∧ abs(j1) < 2 ∧ i0 ≤ 0 ∧ 0 < 2 ∧ i - 1 < 0, i0 >> (abs(j) - 1) = (i0 + 1 - 2 * 2 ^ (abs(j) - 2)) / (2 * 2 ^ (abs(j) - 2)), 2 * 2 ^ (abs(j) - 2) + i < 0, 1 < abs(j), i < 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) ⊦ i0 >> (abs(j) - 1) = (i + 1 - 4 * 2 ^ (abs(j) - 2)) / (4 * 2 ^ (abs(j) - 2)), j = 0, abs(j) = 1

Rule argument

seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i - 1, 2, i0, j1] rotate: T spec: int-div inst: name: div-elim-negpos

Additional information

Used simplifier rules: ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;%⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0); (elim) Used terms: i - 1, 2, i0, j1