Truth Trees, Propositional Logic where conclusion is not related to premises I have a problem that involves an argument in Propositional Logic. However, the conclusion has nothing to do with the premises (completely different variables). I'm fairly certain that this makes the argument invalid, since the conclusion does not necessarily follow from the premises, but I have no idea how to show this/describe this by means of a truth tree (Required for a homework assignment).
Google searching for similar such situations did not yield any guidance, so I was hoping I could get some direction here. The situation is similar to the one below (Since this is homework, I won't give the actual problem; I just need a direction to go in).
$$A \land (B \lor C)$$
$$D \implies K$$
$$(F \iff C) \implies \lnot C$$
$$\lnot K$$
THEREFORE:
$$P \land Q$$  
Not sure how to make a Truth Tree relevant here. Thank you!
 A: Amplifying some of the comments: The basic principle of testing an argument 
$$A_1, A_2, \ldots A_n \therefore C$$
for classical validity using a truth tree is to start the trunk of an (unsigned) tree with
$$A_1\\ A_2\\ \ldots \\ A_n\\ \neg C$$
apply the tree building rules, and if (and only if) every branch of a completed tree closes, then the argument is valid.
Note there is nothing in that which says that every wff on the trunk of the tree has to get used in the tree-building. That's why if the original argument comes out as valid by the tree test, so does
$$A_1, A_2, \ldots A_n, B \therefore C$$
with any irrelevant additional premiss you like. 
For another cases where there we can have a tree closing without invoking every wff of the trunk, suppose the tree starts
$$P\\ \neg P\\ \neg C$$
Then instantly in closes (whatever $C$ is) just in virtue of the initial contradictory pair. But this shows that 
$$P, \neg P\therefore C$$
is classically valid. Similarly, it is easy to see that e.g. the tree starting
$$P\\ P \to Q\\ \neg Q\\ \neg R$$ closes (just using the first three items). And that shows
$$P, P \to Q, \neg Q\therefore R$$
is classically valid. 
For more on this, see any standard text that uses trees (Wilfrid Hodges's, mine, or -- freely available online -- Paul Teller's)
A: We can show that the argument is not valid trying to find a truth assignment $v$ to all sentential letters such that all the premises are true with $v$ while for the conclusion we have : $v(P \land Q)=F$.
As you said, due to the fact that nor $P$ nor $Q$ are used in the premises, we may suspect that we will succeed: the only care is about the potential unsatisfiability of the set of premises.
If by chance the four premsies are unsatisfiable, then it is vacuously true that they logically imply the conclusion.
Formally :


if no truth assignment satisfies every member of a set of formulae $\Sigma$, then for any formula $\alpha$, it is vacuously true that $\Sigma \vDash \alpha$.



For a formal proof with truth tree method I'll try with a "simplified" approach.
In order to show that the argument is valid, we have to show that the four premises and the negation of the conclusion are unsatisfiable.
Thus, we have to start with :
1) $A∧(B∨C)$
2) $D \rightarrow K$
3) $[(F \rightarrow C) \land (C \rightarrow F)] \rightarrow ¬C$
4) $¬K$
5) $\lnot (P∧Q)$

i) get rid of 2) branching into : $i_L) : \lnot D$ and $i_R) : K$
ii) use 4) for closing $i_R)$; thus we have the left branch only, with : $\lnot D$ and $\lnot K$
iii) use 1) to extend the branch to : $\lnot D, \lnot K, A, B \lor C$
iv) now we use $B \lor C$ to branch again into : $iv_L) : \lnot D, \lnot K, A, B$ and $iv_R) : \lnot D, K, A, C$
v) use 5) to branch again with $\lnot P$ and $\lnot Q$ respectively, and consider only the leftmost branch : $v_L) : \lnot D, \lnot K, A, B, \lnot P$
vi) only 3) is left and we have to branch all four previous paths with $\lnot [(F \rightarrow C) \land (C \rightarrow F)]$ and $\lnot C$ respectively
vii) consider now the right branch obtained from $v_L)$ by adding $\lnot C$. It is :

$\lnot D, \lnot K, A, B, \lnot P, \lnot C$.

On this path there are no more formulae to use, thus the path is finished.
Obviously it is satisfiable, by a truth assignment $v$ such that :

$v(A)=v(B)=T$ and $v(C)=v(D)=v(K)=v(P)=F$.

The values of $v$ for $F$ and $Q$ are undefined.
Having found a finished path that does not close (i.e. not "crossed") the five formulae are satisfiable; thus having found an interpretation such that all four premises are true while the conclusion is false, we may conclude that the premises do not logically imply the conclusion.

We can check if the proof is correct in this way.
Extend the truth assignment $v$ by : $v(F)=v(Q)=T$.
Thus we have :
5') $v(P \land Q)= F \land T = F$
4) $v(¬K)=T$
3) $v([(F \rightarrow C) \land (C \rightarrow F)] \rightarrow ¬C) = v([(T \rightarrow F) \land (F \rightarrow T)] \rightarrow T)=v((F \land T) \rightarrow T)=v(F \rightarrow T)=T$
2) $v(D \rightarrow K)=v(F \rightarrow F)=T$
1) $v(A∧(B∨C))=v(T \land (T \lor F))=v(T \land T)=T$.
It's done !
