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Proof step 11 from proof rightshift-8 (previous step)

0 ≤ i ⊦ # (¬ 128 ≤ i ⊃ i2b(i) ';(¬ 256 ≤ i ⊃ i2b(0) ' + i2b(i - 256) ';int2bytes(i / 256) + i2b(i % 256) ')) = 1, i / 256 = bytes2int((¬ 128 ≤ i ⊃ i2b(i) ';(¬ 256 ≤ i ⊃ i2b(0) ' + i2b(i - 256) ';int2bytes(i / 256) + i2b(i % 256) ')).butlast)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ # by ' = 0 +1;