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Proof step 69 from proof rightshift-8-01 (previous step)

j = 1, i % 256 = 0, i→n(j) < 2, i < ~128, ~256 < i, ~256 ≤ 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))) ⊦ 0 = bytes2int(firstn(2 - i→n(j), i2b(~1) ' + i2b(i) ')), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
i % 256 = 0, i < ~128, ~256 < i, ~256 ≤ i, i < 0, ∀ i, j0. i→n(j0) < i→n(1) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);%~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ firstn(1, by ' + bytes) = by ';