Exclusive Or -Logic Find a formula using only the connectives: conjunction, disjunction, and negation that is equivalent to,   P exclusive or Q. I cannot figure out a way to come up with a logical equivalent statement without randomly guessing.
 A: You want $p$ to be true or $q$ to be true, but not both at the same time, thus all of the below formula would give you the correct result:


*

*"$p$ is true or $q$ is true, and it's not the case that both $p$ and $q$ are true"
$(p \lor q) \land \neg (p \land q)\\$

*"$p$ is true and $q$ is false, or $p$ is false and $q$ is true1"
$(p \land \neg q) \lor (\neg p \land q)\\$

*(explaining @Jon Mark Perry's first forumla) "it is neither the case that both $p$ and $q$ are true nor that both $p$ and $q$ are false2"
$\neg ((p \land q) \lor (\neg p \land \neg q))\\$

*(explaining Jon Mark Perry's second one) "$p$ is true or $q$ is true, but either $p$ or $q$ must be false3"
$(p \lor q) \land (\neg p \lor \neg q)\\$
Of course, there is an infinite number of logically equivalent statements (e.g. $(a \land a)$ would trivially be another possible formula to just saying $(a)$, now replace $(a)$ by one of the inclusive-or-formulas and you got another one), but the above listed should be the most intuitive ones.
1 By the inclusive logical or $\lor$ used here, the statement would also be true if both sides of the formula were true, but that will never happen anyway, since $p \land \neg p$ is a contradiction.
2 $\neg(a \lor b)$ ("not $a$ or $b$") reads as "neither $a$ nor $b$".
3 Again, the inclusive or in the second part of the statement would also allow for both $p$ and $q$ to be false, but that possibility is ruled out be the first part of the conjuntion.
A: We have:
$$\lnot ((P \land Q) \lor (\lnot P \land \lnot Q))$$
and:
$$(P\lor Q)\land (\lnot P \lor \lnot Q)$$
From the comments:
$$(P\lor Q) \land \lnot(P \land Q)$$
addendum: adding lemontree's for completeness:
$$(P \land \lnot Q) \lor (\lnot P \land Q))$$
A: Given a truth table of a boolean function, you can always create a boolean function that is equivalent using only conjunction, disjunction, and negation.
If one looks at the truth value of xor, it is only true in two cases: if $p$ is true and $q$ is false OR $p$ is false and $q$ is true which can be written as $(p \wedge \overline{q}) \vee (\overline{p} \wedge q)$.
