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Proof step 6 from proof rightshift-8-01 (previous step)

i→n(j) < # int2bytes(i), ∀ 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))) ⊦ i >> 8 * abs(j) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ 0 ≤ i ∧ 8 ≤ j → i >> j = i / 256 >> (j - 8)
Proof step 8:
¬(0 ≤ i ∧ 8 ≤ 8 * abs(j)), i→n(j) < # int2bytes(i), ∀ 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))) ⊦ i >> 8 * abs(j) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))
Proof step 94:
0 ≤ i ∧ 8 ≤ 8 * abs(j), i→n(j) < # int2bytes(i), ∀ 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))) ⊦ i / 256 >> (8 * abs(j) - 8) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))

Rule argument

seq: ⊦ 0 ≤ i ∧ 8 ≤ j → i >> j = i / 256 >> (j - 8) subst: [j, i] → [8 * abs(j), i] rotate: F spec: intops inst: name: shiftright-rec-8

Additional information

Used terms: 8 * abs(j), i