How to make truth table I have
$F(p, q) = (p \land q) \lor (\lnot p \land q)$
and i need to prove that
$\lnot F(\lnot p, \lnot q) \Leftrightarrow F(p, q)$
is not tautology.
I made truth table for $F(p, q)$ and it looks like this:
$$ 
\begin{array}{c||c||c}
p & q & F(p,q)\\
\hline
\top & \top & \top\\
\hline
\top & \bot & \bot\\
\hline
\bot & \top &\top\\
\hline
\bot & \bot &\top\\
\end{array} 
$$
After simplifying $\lnot F(\lnot p, \lnot q)$ i got
$\lnot F(\lnot p, \lnot q) = (p \lor q) \land (\lnot p \lor q)$ .
Now, I am not sure how to make truth table for this and how to compare these two.
 A: Try filling in the rest of this truth table:
\begin{array}{c|c|c|c|c|c}
p & q & F(p,q) & \lnot p &\lnot q& F(\lnot p,\lnot q)&\lnot F(\lnot p,\lnot q)\\
\hline
\top & \top & \top & \bot & \bot \\
\hline
\top & \bot & \bot\\
\hline
\bot & \top &\top\\
\hline
\bot & \bot &\bot\\
\end{array}
You can easily find the values of $F(\lnot p,\lnot q)$
by using the first three columns of the table.
For example, on the first line of the table $\lnot p = \bot$ and
$\lnot q = \bot$, and you find the value of $F(\bot,\bot)$ on
the last line of the table.
I find that if there is any amount of complication in the
logical expressions, it almost always makes truth-table problems much easier
if I insert auxiliary columns like this to show the intermediate steps
of the calculations.  Of course you still want to compare only the columns 
for $F(p, q)$ and $\lnot F(\lnot p,\lnot q)$ in the end.
As you may discover, there is really no need to "simplify"
$\lnot F(\lnot p,\lnot q)$ if you use the truth-table entries
you already discovered for $F(p, q)$ in this manner.

Edit: As pointed out by Mauro ALLEGRANZA, the entry for $F(p,q)$ on
the last line was incorrect in the truth table originally presented in
this answer (as well as in the truth table in the original question,
from which that entry was copied). This does not invalidate the method
described in this answer, but it does have an effect on the values
that result on the second line of the table.
A: You found out that $\neg F(\neg p, \neg q) = (p \lor q) \land (\neg p \lor q)$, so now simply extend your truth table:
$$ 
\begin{array}{c||c||c||c}
p & q & F(p, q) = (p \land q) \lor (\lnot p \land q) & \neg F(\neg p, \neg q) = (p \lor q) \land (\neg p \lor q)\\
\hline
\top & \top & \top & ?\\
\hline
\top & \bot & \bot & ?\\
\hline
\bot & \top &\top & ?\\
\hline
\bot & \bot &\top & ?\\
\end{array} 
$$
If the two columns $F(p,q)$ and $\neg F(\neg p, \neg q)$ have the same entries, then $F(p,q) \iff \neg F (\neg p, \neg q)$ is a tautology, otherwise it is not.
