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Proof step 151 from proof shiftright-rec-8-neg (previous step)

(i1 * 256 + j1 - 1) >> (j - 1) = ((j1 - 1) / 256 + i1 - 1) >> (j - 9), 256 ≤ abs(j1 - 1), i1 < 0, i1 * 256 + j1 < 0, i1 < j, j1 < j, 8 < j, 0 < j, i1 * 256 + j1 < j, ~256 < j1, j1 < 0, 0 < abs(j1 - 1) + j1, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ (i1 * 256 + j1 - 1) >> (j - 1) = (i1 - 1) >> (j - 9), (j1 - 1) % 256 = 0, j = 8, ((i1 * 256 + j1) * 2 - 1) % 256 = 0, (j1 - 1) / 256 = 0, j = 0, j1 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j1 - 1] rotate: F spec: int-abs inst: name: Abs-neg

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 < j → ¬ j ≤ i;% ⊦ ¬ j < 0 → abs(j) = j;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 - i = ~ i;% ⊦ ~(i - 1) = ~ i + 1;% ⊦ i + ~ j = i - j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;%i + 1 = i0 ⊦ (* i) < j → (j ≤ (* i0) ↔ j = i0);% ⊦ i + (j - i) = j;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i - i1 - j = i - i1 + j;% ⊦ i * j - i = i * (j - 1);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used terms: j1 - 1