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Proof step 56 from proof rightshift-8-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 57:
abs(i) < abs(j) ⊦ i / j = 0
Proof step 58:
¬ abs(i) < abs(256), i % 256 = 0, i→n(j) < # int2bytes(i), i < 0, ~256 ≤ 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))) ⊦ j = 0, i / 256 >> 8 * (abs(j) - 1) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))
Proof step 61:
abs(i) < abs(256), i % 256 = 0, i→n(j) < # int2bytes(i), i < 0, ~256 ≤ 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))) ⊦ j = 0, 0 >> 8 * (abs(j) - 1) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))

Rule argument

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

Additional information

Used terms: 256, i