How to prove if something is false or true? I'm a bit stuck at a task i'm working on here.
Here is the task:
Show statement "If $(P \rightarrow Q)$ and $(Q \rightarrow R)$ is true, then $(P \rightarrow R)$ is true by a counter positive-evidence
This is what i've done so far: 
Since the task asks for counter positve-evidence I changed the statement to:
"If $(\lnot P \rightarrow \lnot Q)$ and $(\lnot Q \rightarrow \lnot R)$ is false, then $(\lnot P \rightarrow \lnot R)$ is false.
The truth table for this statement should look like this:

How can I see if the counter positive-evidence is wether true or false? Or have I done something wrong here?
Thanks alot for help!
 A: Assuming you mean that you need to prove the contra-positive, recall that for any implication $$ p\rightarrow q \equiv \underbrace{\lnot q \rightarrow \lnot p}_{\text{contrapositive}}$$
So the contrapositive of $A$ you're needing to prove, where
$$A\equiv \text{If } (P→Q)\text{ and } (Q→R) \text{ is true,  then } (P→R)\text{ is true"}$$ is given by the proposition $B$: $$B\equiv \text{ if} \;\lnot (P \rightarrow R)\;\text{ then } \lnot[(P\rightarrow Q)\;\text{and}\; (Q\rightarrow R)]$$ which is the proposition $$\lnot (P \rightarrow R) \rightarrow \lnot[(P \rightarrow Q) \land (Q \rightarrow P)]$$
That is, you need to establish that "if $P\rightarrow Q$ is false, then $[(P\rightarrow Q) \land (Q \rightarrow R)]$ must be false."
Be careful, too, when negating an implication:
$$\lnot (P \rightarrow Q) \not\equiv (\lnot P \rightarrow \lnot Q)$$
Rather $$\lnot (P \rightarrow Q) \equiv \lnot(\lnot P \lor Q) \equiv P \land \lnot Q$$
We'll show that the contrapositive of your statement is indeed true, and hence the statement is true:

The truth table above, generated by Wolfram Alpha, gives evidence that the implication (the contrapositive itself) is true no matter what the truth values of $P, Q, R$: it is always true and hence is a tautology. 
When constructing truth tables, it helps to do the evaluation in steps, with "sub-formulas" from the entire proposition, for example, including the antecedent and the subsequent, and evaluating each, etc. Then use the rules for the connectives to evaluate the sub-formulas, and finally, using the rule for the MAIN connective (here $\rightarrow$), evaluate the entire proposition.
A: Note that $P \to Q$ is equivalent to $\lnot P\vee Q$, you can verify this with a truth table. Then the negation is
$$ \lnot(P \to Q)= \lnot (\lnot P \vee Q)= \lnot(\lnot P) \wedge \lnot Q = P \wedge \lnot Q.$$
