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Proof step 137 from proof shiftright-rec-8-neg (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 138:
abs(i) < abs(j) ⊦ i / j = 0
Proof step 139:
¬ abs(j1 - 1) < abs(256), (j1 - 1) >> (j - 1) = ((j1 - 1) / 256 - 1) >> (j - 9), 8 < j, 0 < j, j1 < j, ~256 < j1, j1 < 0, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 0, (j1 - 1) >> (j - 1) = ~1 >> (j - 9), (j1 - 1) % 256 = 0, j = 8, (2 * j1 - 1) % 256 = 0, (j1 - 1) / 256 = 0

Rule argument

seq: abs(i) < abs(j) ⊦ i / j = 0 subst: [j, i] → [256, j1 - 1] rotate: F spec: int-div inst: name: div-zero-02

Additional information

Used terms: 256, j1 - 1