Explanation:
The required combinatorial circuit can be constructed as follows:
Firstly, we need to find the negation of p, q and r which can be achieved using inverters as shown below:
p q r
| | |
| | |
NOT NOT NOT
| | |
| | |
~p ~q ~r
Next, we can use OR gates to form the term (¬p ∨ ¬r) and (q ∨ r) as shown below:
~p ~r q r
| | | |
| | | |
OR OR | |
| | | |
| | | |
pORr rORp qORr rORq
Then, we can use AND gates to form the terms (¬p ∨ ¬r) ∧ ¬q and ¬p ∧ (q ∨ r) as shown below:
~p ~r q r
| | | |
| | | |
OR OR | |
| | | |
| | | |
pORr rORp qORr rORq
| | | |
| | | |
NOT NOT NOT |
| | | |
| | | |
~q ~pORr ~qORr qORr
| | | |
| | | |
AND OR AND |
| | | |
| | | |
(~pORr ~q) qORr qORr
| | |
| | |
OR AND |
| | |
| | |
(~pORr ∨ ~q) (~pORr ∧ qORr)
| | |
| | |
| OR |
| | |
| | |
((~pORr ∨ ~q) ∨ (~pORr ∧ qORr))
| | |
| | |
| | |
OUTPUT
((~pORr ∨ ~q) ∨ (~pORr ∧ qORr))