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Proof step 3 from proof rightshift-8-01-01 (previous step)

¬ i→n(j) < # int2bytes(i) ⊦ i >> 8 * j = bytes2int(firstn(# int2bytes(i), fillsgn(i→n(j) + # int2bytes(i), int2bytes(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# int2bytes(i) ≤ i→n(j) ⊦ i >> 8 * j = bytes2int(firstn(# int2bytes(i), fillsgn(i→n(j) + # int2bytes(i), int2bytes(i))))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;