Show that conditional statement is a tautology without using

Show that conditional statement is a tautology, without using truth table. [(p rightarrow q)^(q rightarrow r)] rightarrow (p rightarrow r)

Solution

[(p->q) (q->r )] ->(p->r)

=[~p v q ~q v r ]->(p->r).............[ q ~q=F]

=[~p v F v r]->(p->r)

=F->(p->r)   [F-> anything =T]

=T