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Proof step 15 from proof rightshift-8-01 (previous step)

i→n(j) < # int2bytes(i), i < 0, ∀ i, j0. i→n(j0) < i→n(j) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ j = 0, (i - 255) / 256 >> 8 * (abs(j) - 1) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i))), i % 256 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ i % j ≠ 0 ∧ j0 = j - 1 ∧ i < 0 ∧ 0 < j → (i - j0) / j = i / j - 1
Proof step 17:
i % 256 ≠ 0 ∧ 255 = 256 - 1 ∧ i < 0 ∧ 0 < 256, i→n(j) < # int2bytes(i), i < 0, ∀ i, j0. i→n(j0) < i→n(j) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ j = 0, (i / 256 - 1) >> 8 * (abs(j) - 1) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i))), i % 256 = 0

Rule argument

seq: ⊦ i % j ≠ 0 ∧ j0 = j - 1 ∧ i < 0 ∧ 0 < j → (i - j0) / j = i / j - 1 subst: [j, j0, i] → [256, 255, i] rotate: F spec: int-div inst: name: div-minus-onediff

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: 256, 255, i