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Proof step 97 from proof rightshift-8-01 (previous step)

¬ 255 < i, i→n(j) < # int2bytes(i), 0 ≤ 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) - 1) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i))), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 98:
i→n(j) < # int2bytes(i), 0 ≤ i, i ≤ 255, ∀ 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) - 1) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i))), j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;