Need help with tautology proof without truth tables. I am trying to prove $$[(p\to q)~\&~(q\to r)]\to (p\to r) $$ is a tautology using only logical laws. I have gotten part-way there but I got stuck and am not sure how to proceed. Please state any laws that you use in your answers so that I can reference them. ~ = NOT & = AND V = OR -> = IMPLIES The Proof: $$\begin{array}{ll} [(p\to q)~\&~(q\to r)]\to (p\to r); & \text{Given} \\\ \sim[(\sim p\vee q)~\&~({\sim} q\vee r)]~\vee~({\sim} p\vee r); & \text{Material Implication} \\\ [{\sim}({\sim} p~\vee~ q)\vee{\sim}({\sim} q\vee r)]\vee({\sim} p\vee r); & \text{DeMorgan's Law} \\\ [(p~\&\,{\sim} q)\vee(q~\&\,{\sim} r)]\vee({\sim} p\vee r); & \text{DeMorgan's Law} \end{array}$$

First remove redundant brackets because `V` is associative:


(p&~q)V(q&~r)V~pVr


Then rearrange a little since `V` is commutative:


~pV(p&~q)VrV(q&~r)


Then we distribute the first two terms, and we distribute the last two:


[(~pVp)&(~pV~q)]V[(rVq)&(rV~r)]


We can cancel `(~pVp)` and `(rV~r)` because they're both tautologies (I don't know what you call that law) and because `&`-ing with a tautology doesn't change anything (I don't know what you call that law). Then again remove redundant brackets to get


~pV~qVrVq


and you should be able to see why this is a tautology.