Write the logical statement of this circuit - am I correct? 
my ans: $$(p \land q)\lor(\lnot p \land q)$$
What is the simplified version of the correct answer? $p\lor q$
 A: Your initial "answer" is correct, but can be simplified:
Recall the distributive law (DL): $(a\lor b) \land c \equiv (a \land c) \lor (b \land c)$.
And recall the law of the excluded middle (LEM): $p\lor \lnot p\equiv T$ where $T \equiv \;\text{true}\;\equiv 1$.
$$\begin{align}(p\land q)\lor (\lnot p \land q) & \equiv (p\lor \lnot p) \land q\tag{DL} \\ \\
& \equiv T \land q \tag{LEM}\\ \\ 
& \equiv q\end{align}$$
A: You can write down the truth table to figure out the answer, or use De Morgan's and "usual" laws to factorize by $q$: 
$$(p\land q )\lor (\lnot p\land q) = (p \lor \lnot p )\land q = q$$
A: I don't think you've got it quite right.  From what I can see of the diagram--it displays pretty small on my 'droid--it is, in words, $p$ in series with $q$, as a "unit", in parallel with the "unit" $\bar{p}$ in series with $q$.  So this makes your Boolean formula $(p \wedge q) \vee (\bar{p} \wedge q)$ correct.  (I'm using overbar rather than "~" for logical negation because I can't remember the Latex for "~".)  However, we have, by the distributive laws for Boolean logic, 
$(p \wedge q) \vee (\bar{p} \wedge q) = ((p \vee \bar{p}) \wedge q)$,
and since $p \vee \bar{p} = 1$, this becomes
$1 \wedge q = q$,
the final answer.  This makes sense since one of $p$, $\bar{p}$ will always be "on".  
Hope this helps.  Cheers.
A: Hint: Without even considering any equations or logical statements, look at the circuit. As long as $q$ is closed, does it matter if $p$ is true or false?
