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Proof step 21 from proof shiftright-rec-8-neg (previous step)

i % 256 = 0, ~128 ≤ i, i < 0, ∀ i, j0. i→n(j0) < i→n(8) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)

Rule argument

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

Additional information

Used simplifier rules: ⊦ 0 < i ∧ j < 0 → ((* j) ≤ (* i) * i0 ↔ j / i ≤ i0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used terms: 256, i