Show that (P→Q) ∨(Q→R) is a tautology I don't really understand Tautologies or how to prove them, so if someone could help, that would be great! 
 A: Truth table method:
$$\matrix{
  p&q&r&p\to q&q\to r&\hbox{answer}\cr
  T&T&T&T&T&T\cr
  T&T&F&T&F&T\cr}$$
and so on.  You can provide six more rows yourself and check that the final answer is always true.
Logical equivalences method:
$$\eqalign{
  (p\to q)\vee(q\to r)\quad
  &\Leftrightarrow\quad (\neg p\vee q)\vee(\neg q\vee r)\cr
  &\Leftrightarrow\quad (q\vee\neg q)\vee (\neg p)\vee r\cr
  &\Leftrightarrow\quad {\bf T}\vee (\neg p)\vee r\cr
  &\Leftrightarrow\quad {\bf T}\ .\cr}$$
Saving the best to last. . . smart idea method: $q$ is either true or false.  If $q$ is true then $p\to q$ is true; if $q$ is false then $q\to r$ is true.  In either case, $(p\to q)\vee(q\to r)$ is true.
A: If $Q$ is true, then $P\to Q$ is true, and we're done.
If $Q$ is false, the $Q\to R$ is true, and we're done.
But to go back to basic definitions, a tautology is something that is true in every row of the truth table.  In this case there are eight rows.  Look at each row and ascertain whether the proposition in question is true.
\begin{array}{ccc}
P & Q & R \\
\hline
T & T & T \\
T & T & f \\
T & f & T \\
f & T & T \\
T & f & f \\
f & T & f \\
f & f & T \\
f & f & f
\end{array}
A: If Q is true, the first implication is true, so the disjunction is true.
If Q is false, the second implication is true, so the disjunction is true.
In either case, the disjunction is true, quod erat demonstrandum.
