Show that p q q r pr without using truth table as in dis
Show that [ (p -> q) ^ (q ->r)] -> (p->r) without using truth table. as in discrete math 7th edition chap 1.3 , 12 (b)
Solution
(pq)(qr)(pr)(pq)(qr)(pr)
(¬pq)(¬qr)(pr)(¬pq)(¬qr)(pr)
((¬pq)¬q)((¬pq)r)(pr)((¬pq)¬q)((¬pq)r)(pr)
((¬p¬q)(q¬q))((¬pr)(qr))(pr)((¬p¬q)(q¬q))((¬pr)(qr))(pr)
(¬p¬q)((¬pr)(qr))(pr)
![Show that [ (p -> q) ^ (q ->r)] -> (p->r) without using truth table. as in discrete math 7th edition chap 1.3 , 12 (b)Solution (pq)(qr)(pr)(pq)(qr)( Show that [ (p -> q) ^ (q ->r)] -> (p->r) without using truth table. as in discrete math 7th edition chap 1.3 , 12 (b)Solution (pq)(qr)(pr)(pq)(qr)(](/WebImages/23/show-that-p-q-q-r-pr-without-using-truth-table-as-in-dis-1057773-1761552114-0.webp)