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Proof step 35 from proof rightshift-8 (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

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