Negate and Simplify $p\wedge (q\vee r)\wedge(\neg p\vee\neg q\vee r)$ So far I have
\begin{align*}
\neg[p\wedge(q\vee r)\wedge(\neg p\vee\neg q\vee r)]&\Longleftrightarrow\neg p\vee\neg(q\vee r)\vee\neg(\neg p\vee\neg q\vee r)&\mbox{Demorgan's Law}\\
&\Longleftrightarrow\neg p\vee(\neg q\wedge\neg r)\vee(\neg\neg p\wedge\neg\neg q\wedge\neg r)&\mbox{Demorgan's Law}\\
&\Longleftrightarrow\neg p\vee(\neg q\wedge\neg r)\vee(p\wedge q\wedge\neg r)&\mbox{Double Negation}\\
\end{align*}
At this point I'm unsure of where to go. I considered applying the Distributive Law which would give me
$$[(\neg p\vee\neg q)\wedge(\neg p\vee\neg r)]\vee(p\wedge q\wedge \neg r)$$
But I don't know if this is getting any closer to 'simplifying' the statement.
I have an obvious intuition about when an algebraic expression  is fully simplified. I assume there is an analogous intuition around logical statements but I certainly don't have it. Is there a rule of thumb for this sort of thing? What would the negation of this statement look like 'fully simplified'?
Thanks
 A: In a situation like this, with so few variables (3), and such complex expressions,
the easiest approach is to dispense with all attempts at elegance and simply
construct the appropriate chart of Boolean true/false values.
Let $A$ denote $(q \vee r).$ 
Let $B$ denote $(\neg p \vee \neg q \vee r).$ 
Then you are trying to negate and simplify $(p \wedge A \wedge B).$
$$\begin{array}
{|c|c|c|c|c|c|c|} 
\hline
p & q & r & A & B & (p \wedge A \wedge B) & \neg (p \wedge A \wedge B) \\
\hline
t & t & t & t & t & t & f \\
\hline
t & t & f & t & f & f & t \\
\hline
t & f & t & t & t & t & f \\
\hline
t & f & f & f & t & f & t \\
\hline
f & t & t & t & t & f & t \\
\hline
f & t & f & t & t & f & t \\
\hline
f & f & t & t & t & f & t \\
\hline
f & f & f & f & t & f & t \\
\hline
\end{array}$$
Assuming that I have made no mistake, the "negate and simplify"
will always be true except for the 
$(p,q,r)$ truth/false values of $(t,t,t)$ and $(t,f,t)$ which 
simplifies to $(\neg p \vee \neg r) = \neg(p \wedge r).$
A: Using that (Distributive law and "Glue" law):
$$ a \wedge \left(b\vee c\right) = \left(a \wedge b\right)\vee\left(a \wedge c\right)$$
$$ (a\vee b) \wedge (\neg a\vee b) = b$$
We have:
$$ p\wedge (q\vee r)\wedge(\neg p\vee(\neg q\vee r)) = p\wedge \left(\left((q\vee r) \wedge \neg p\right) \vee \left((q\vee r)\wedge (\neg q \vee r)\right)\right) = $$
$$ = p\wedge \left(((q\vee r) \wedge \neg p) \vee r\right) $$
Using Distributive law again we have:
$$ p\wedge \left(((q\vee r) \wedge \neg p) \vee r\right) = p\wedge \left(((q\vee r) \wedge \neg p) \vee r\right) = (p \wedge (q \vee r) \wedge \neg p) \vee (p \wedge r) $$
Also, we know that:
$$ a \wedge b \wedge \neg a = 0 $$
Now we have:
$$ (p \wedge (q \vee r) \wedge \neg p) \vee (p \wedge r) = 0 \vee (p \wedge r) = p \wedge r$$
The answer is:
$$ p\wedge (q\vee r)\wedge(\neg p\vee\neg q\vee r) = p \wedge r $$
