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Proof step 7 from proof rightshift-8 (previous step)

0 ≤ i ∧ 8 ≤ 8, 0 ≤ i ⊦ i / 256 >> (8 - 8) = bytes2int(int2bytes(i).butlast), # int2bytes(i) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ i ⊦ # int2bytes(i) = 1, i / 256 = bytes2int(int2bytes(i).butlast)

Rule argument

Additional information

Used simplifier rules: ⊦ i >> 0 = i;