Predicate Logic I'm studying for an exam, and I'm not really sure how to portray this. The domain is all people.
$V (w) = w$ is a voter
$P (w) = w$ is a politician
$K (y, z) = y$ knows $z$
$T (y, z) = y$ trusts $z$
Cal ($c$) is a voter who knows everyone. Would this be: $$\forall x V(c)\land K(c,x)$$
There is a politician that no other politician trusts:
$$\exists x\forall y P(x)\land P(y)\land T(y,x)$$
I'm not sure if those are right. Wouldn't the last one be saying, There are politicians that no one trusts? How do I make it singular?
Also: No one trusts every politician:
$$\exists x \forall y P(y)\land T(\lnot x,y)$$
 A: Translation of ordinary language assertions into logical language is sometimes difficult.  Certain subtleties and connotations simply do not translate. There are always a number of logically equivalent translations.  And in a fair number of cases, there can be two logically inequivalent translations that are each sensible. 
Question 1:  This is right.  There are other right answers, such as $V(c) \land \forall x K(c,x)$.
Question 2:  As usual there is more than one translation. We want to say something closer to $\exists x (P(x) \land \forall y((P(y)\land \lnot(y=x))\implies \lnot T(y,x)))$.  
The implication symbol can be avoided in the usual ways, since $A \implies B$ is logically equivalent to $\lnot A \lor B$ or $\lnot(A \land \lnot B)$.
Your version does not contain the negation, and is in many other ways not close. It seems to say among other things that everybody ("$y$") is a politician.  Also, the sentence needs to say that this bad politician $x$ is not trusted by any other politician. So we need to make sure, by using the $\lnot(y=x)$, or in some other way, that we do not claim that this bad politician $x$ does not trust herself.  This question is harder than the other two.
Question 3: The sentence $\forall x  \exists y (P(y)\land \lnot T(x,y))$ should work.  Read this as "for any person $x$, there is a politician $y$ whom $x$ does not trust."
Your version is very far from saying the desired thing. Somehow you have the $\lnot$ inside the $T$ relation.    
There is a case to be made for $\forall y(P(y) \implies \exists x\lnot T(x,y))$.
A: For the second, the untrusted politician is the one that exists and should be the second argument of T.  Also the only ones that we know don't trust him are politicians themselves, so $\exists x \forall y P(x) \land [P(y) \implies \lnot T(y,x)]$.  This asserts that there is at least one politician that no other politician trusts.  There could be more than one.
For the third, you want the quantifiers in the other order, as your are saying for all politicians there exists (someone who may depend upon which politician) that distrusts them.
