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Proof step 14 from proof rightshift-8-01 (previous step)

¬(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 - 255) / 256 >> 8 * (abs(j) - 1) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
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

Rule argument

Additional information

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