Concept:-The total number of 1's must be odd including parity bit generated by Odd parity Generator.
Here consider the input to the Odd Parity generator and corrosponding odd parity bit generated by Odd parity Generator circuit.
| 1 |
0 |
x |
1 |
0 |
1 |
y |
Odd Parity Generator O/P bit |
| |
|
0 |
|
|
|
0 |
0 |
| |
|
0 |
|
|
|
1 |
1 |
| |
|
1 |
|
|
|
0 |
1 |
| |
|
1 |
|
|
|
1 |
0 |
The O/P of of Odd parity generator is clearly here $x \oplus y$
now assume $x \oplus y$ = R
o/p of Mux is z'R+z = R+z = $x \oplus y$ + z
Hence Option C is the Ans.
Note:-
z'R + z
(Now use + is distributed over . property
//Invalid in ordinary algebra but supported in boolean Algebra)
= (z'+z).(R+z)
=1.(R+z)
=R+z