Simplify, equivalent for (p ∨ ¬q) ∧ (¬p ∨ ¬q) In my text book I'm asked to deduce a simpler expression for
(p ∨ ¬q) ∧ (¬p ∨ ¬q)
Looking at an equivalency table I did, it seems p ∨ ¬q gives the same results as (p ∨ ¬q) ∧ (¬p ∨ ¬q). However I'm not sure how you would deduce this without the table, as in, if I was outrightly asked to write the above in simpler terms I wouldn't know where to begin. Am I understanding this correctly?
My Table:

 A: This is actually an instance of the following logical equivalence principle:
Adjacency
$(p \land q) \lor (p \land \neg q) \Leftrightarrow p$
So, with this principle, you can immediately say that:
$(p \lor \neg q) \land (\neg p \lor \neg q) \Leftrightarrow \neg q$
When you do boolean algebra to simplify expressions, this situation comes up a lot, so I highly recommend remembering this equivalence principle!
A: $$(p \lor \lnot q) \land (\lnot p \lor \lnot q) \iff (p \land \lnot p) \lor \lnot q \iff \lnot q.$$
A: Just using the following properties of boolean algebra,
\begin{align}
 a\bar{a} &=0\\
 1 a &=a\\
 0 a &=0\\
 a+\bar{a}&=1\\
 0+a &=a\\
1+a &=1\\
a+a&=a\\
a(b+c)&=ab+ac
\end{align}
you can simplify 
\begin{align}
(p+\bar{q})(\bar{p}+\bar{q})
 &=p\bar{p}+ (p+\bar{p})\bar{q}+\bar{q}\bar{q}\\
 &= 0 + 1\bar{q}+\bar{q}\\
 &= 0 +\bar{q}\\
 &=\bar{q}
\end{align}
A: Here's the corrected table:
$$\begin{array}{cc|cccc|c}
  P & Q & \neg P & \neg Q & P\lor\neg Q & \neg P\lor\neg Q & (P\lor \neg Q)\land(\neg P\lor \neg Q)\\\hline
 T&T&F&F&T&F&F\\
 T&F&F&T&T&T&T\\
 F&T&T&F&F&T&F\\
 F&F&T&T&T&T&T
\end{array}$$
From this, you can see that $(p \lor \neg q)\land(\neg p \lor \neg q) \iff (\neg q)$ 
A: Note that if $q$ is true one of the conjuncts will be true and the other false, and the conjunction will be false, whereas if $q$ is false both conjunts will be true, so the conjunction will be true. Since it's true iff $q$ isn't, the simplest expression for the truth-function is $\neg q$.
A: We can do this using propositional logic.  With some mild short-cuts for brevity, we have:
1.  Assume: (p ∨ ¬q) ∧ (¬p ∨ ¬q)
2.  Conclude: p ∨ ¬q (from 1)
3.  Conclude: ¬p ∨ ¬q (from 1)
4.  Suppose p:
5.      Conclude ¬q (from 2,4)
6.  Conclude p → ¬q (from 4-5)
7.  Conclude ¬q ∨ ¬q (from 2, 6)
8.  Conclude ¬q (from 7)

I'll note that it is common to conclude 6 directly from 2.
So, this shows that ((p ∨ ¬q) ∧ (¬p ∨ ¬q)) → ¬q .  However, now we have to ask: Does it show anything else?  Let's apply basic substitution to check:
(p ∨ ¬q) ∧ (¬p ∨ ¬q)
(p ∨ ⊤) ∧ (¬p ∨ ⊤) (We know ¬q, so we can learn nothing further about q.  Replace ¬q with T.
⊤ ∧ T (Anything ∨ T) is always true, regardless of the value anything.  So that tells us nothing.  
So, we can now conclude:
p ∨ ¬p (This is a tautology due to law of excluded middle, so it can remain unstated)
¬q
An even simpler proof would be proof by contradiction (i.e., assume q, then get (p) ∧ (¬p)).  This is the approach taken by J.G.
