How to solve a statement with contradiction evidence? I'm trying to solve the statement below with contradiction evidence.

If $(P \rightarrow Q)$ and $(Q \rightarrow R)$ is true, then $(P
 \rightarrow R)$ is true.

This is what i've done so far:

$((P \rightarrow Q)$ $\land$ $(Q \rightarrow R))$ $\rightarrow$  $(P \rightarrow R)$

This is the truth table I created for the statement:

The truth table shows us that the statement is true since all the valuations in 
$((P \rightarrow Q)$ $\land$ $(Q \rightarrow R))$ $\rightarrow$  $(P \rightarrow R)$ is true.
I'm I doing this the correct way or is there something i'm missing?
Thanks in advance!
 A: Whe does $(P \rightarrow Q) \land (Q \rightarrow R)\rightarrow (P \rightarrow R)$ evaluate to $0$?
$$(P \rightarrow Q) \land (Q \rightarrow R)\rightarrow (P \rightarrow R) \equiv 0   \tag{1}$$
if and only if
  $$(P \rightarrow Q) \land (Q \rightarrow R) \equiv  1  \tag{2} $$
and
  $$P \rightarrow R \equiv  0 \tag{3}$$
Now $(2)$ implies
$$ P \rightarrow Q \equiv 1 \tag{4}$$
and
$$Q \rightarrow R \equiv  1 \tag{5}$$ 
From $(3)$ we can conclude
$$P \equiv  1 \tag{6}$$
$$ R \equiv  0 \tag{7}$$
$(5)$ and $(7)$ gives
$$Q \equiv 0 \tag{8}$$
$(4)$ and $(8)$ gives
$$P \equiv 0 \tag{9}$$
but $(9)$ contradicts $(6)$. So there are no truth values for $P$, $Q$ and $R$ such that $1$ is valid.
A: Your proof directly proves the given statement. However, if the task here is to prove the statement by "contradiction", we take the premises $P \rightarrow Q$ and $Q\rightarrow R$, and also assume the negation of the conclusion, here, $\lnot(P \rightarrow R)$. Then the aim is to derive a contradiction. Upon deriving a contradiction given the assumtion, we conclude then that if the premise is true, the conclusion must follow: we will have proven the same implication you proved directly.
Can you construct the truth table for $$(P \rightarrow Q) \land (Q\rightarrow R) \land \lnot (P \rightarrow R)\quad??$$
You should find that the proposition above yields a contradiction, the opposite of a tautology (table compliments of Wolfram Alpha):

and as a contradiction, we now reject the assumption that leads to the contradiction.  We conclude $$\begin{align} \lnot\Big([(P \rightarrow Q) \land (Q \rightarrow R)] \land \lnot (P \rightarrow R)\Big) &\equiv \lnot [(P \rightarrow Q) \land (Q\rightarrow R)] \lor (P \rightarrow R) \\ \\ & \equiv [(P\rightarrow Q) \land (Q\rightarrow R)] \rightarrow (P \rightarrow R)\end{align}$$ That is, we have proven that if the premise $(P\rightarrow Q) \land (Q\rightarrow R)$ holds, then the conclusion $P \rightarrow R$ also holds.
A: This is a fine way to prove that statement.  I don't know what "with contradiction evidence" means, so can't say whether you have met that requirement.  It sounds like you are supposed to start from premises $P \rightarrow Q$ and $Q \rightarrow R$, then assume $\lnot (P \rightarrow R)$ and derive a contradiction.
