Simplify Logic using One Connective: Solving ((q→p) ^ (p→r))→(r→q)

[c4nn3t]

So I've got

((q→p) ^ (p→r))→(r→q)

And I have to simplify it down as much as possible using only one logical connective for the end result (not one type, just one total). Here's been my workflow so far:

((¬q v p) ^ (¬p v r))→(¬r v q)
¬((¬q v p) ^ (¬p v r)) v (¬r v q)
¬(¬q v p) v ¬( ¬p v r)) v (¬r v q)
(q ^ ¬p) v (p ^ r) v (¬r v q)

From here, I'm not sure how to boil down this DNF form into a single 'x and/or/implies y'

Mucho thanks in advance ;)

 

[Evgeny.Makarov]

(q ^ ¬p) v (p ^ r) v (¬r v q)

The middle disjunct should be $p\land\neg r$.

Use the fact that $(x\land y)\lor x=x$. Indeed,
\[
(x\land y)\lor x=(x\land y)\lor (x\land 1)=x\land (y\lor 1)=x\land 1=x.
\]