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Proof step 4 from proof rightshift-8-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i→n(j) < # int2bytes(i) ⊦ i >> 8 * abs(j) = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ abs(i * j) = abs(i) * abs(j);