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Proof step 70 from proof rightshift-8-01 (previous step)

i % 256 = 0, i < ~128, ~256 < i, ~256 ≤ i, i < 0, ∀ i, j0. i→n(j0) < i→n(1) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 71:
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 → ((* i) * i0 < (* j) ↔ i0 < j / i);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 < i ∧ j < 0 → ((* j) < (* i) * i0 ↔ (j + 1 - i) / i < i0);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: 256, i