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Proof step 76 from proof shiftright-rec-8-neg (previous step)

(i - 1) / 2 >> (j - 1) = ((i - 1) / 2 - 255) / 256 >> (j - 9), ((i - 1) / 2) % 256 = 0, 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, ((i - 1) / 2 - 255) / 256 = (i / 256 - 2) / 2, j = 0, (i - 1) / 2 >> (j - 1) = (i / 256 - 2) / 2 >> (j - 9)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 77:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 78:
256 ≠ 0, (i - 1) / 2 >> (j - 1) = ((i - 1) / 2 - 255) / 256 >> (j - 9), ∃ j0. (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, ((i - 1) / 2 - 255) / 256 = (i / 256 - 2) / 2, j = 0, (i - 1) / 2 >> (j - 1) = (i / 256 - 2) / 2 >> (j - 9)

Rule argument

seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [256, (i - 1) / 2] rotate: F spec: int-div inst: name: modex

Additional information

Used terms: 256, (i - 1) / 2