Logical expression representation Let P be a logical expression in two variables, i.e.,
$P:(T, F) × ( T, F ) → ( T, F)$
Express the following using only quantiﬁers, logical operators, parentheses, and the
logical expression P(q, r).
i) P(q, r) is a tautology. 
ii) P(q, r) is a contingency. 
iii) P(q, r) is a contradiction.
For the first part of this question, I get an answer: $\forall q \exists r P(q,r) $
since a tautology is always true. The way I thought of it, is that for any possibility of TRUE or FALSE in the Cartesian product above, there is only one possible combination that would make the "if then" statement a tautology.
The set: {$(T,T),(T,F),(F,T),(F,F)$}
$T \to T \equiv T $
$F \to T \equiv T $
So for any value I choose for $q$ there exists a value $r$ that would make this a tautology. The rest of the combinations all give FALSE values.
The answer to this question is not what I got, but rather: $\forall q \forall r P(q,r) $
How can this be? Is there something I'm doing wrong?
 A: P(q, r) is not an object language expression, but rather a meta-language expression (it's not a logical expression, but a metalogical expression).  It doesn't fall into one of the three categories of qualifying as a "tautology", "contradiction", or "contingency".
The text probably intends P(q, r) to indicate any object logic expression which has two (equiform) variables.  In Polish notation some examples of this class of expressions are: Cpq, CCpqCqp, CCpCqpKCpqp, CKpqKqp, CKpqAqp, CCpqCCpqCqp, CCpqCCppCpq, and CApqCCCCAKpqppqqq.  They all have two variables.   A wff with only two (equiform) variables only holds as a tautology when for all values for the first variable, and for all values of the second variable, the evaluation of the wff always evaluates to the designated value.  Your formula doesn't say something equivalent to that. 
A: You get confused between satisfiability and validity. $\forall q \exists r P(q,r) $ does not guarantee P(q,r) to be tautology since $\exists r \neg P(q,r) $.
