@ Bitovska disjunkcija
@ Ulaz: x (f0), y (f1)
@ Izlaz: z = x | y (f2)

f0 aa 55

@ Pomocni podatak: t = ~x (f3)


00 00f0   @ MUA x
02 51ff   @ XOR #ff  (~x)
04 10f3   @ AUM t    (t = ~x)
06 00f1   @ MUA y 
08 51ff   @ XOR #ff  (~y)
0a 40f3   @ AND t    (~y & ~x)
0c 51ff   @ XOR #ff  ~(~y & ~x) = y | x
0e 10f2   @ AUM z    z = y | x
10 80     @ HALT
