# Thread: is this logical expression equivalent?

1. ## is this logical expression equivalent?

p + q = (pVq) ^ -(p^q)

2. Originally Posted by EquinoX
p + q = (pVq) ^ -(p^q)
what does the + mean? have you tried doing a truth table?

3. it means an exclusive or

4. Originally Posted by EquinoX
it means an exclusive or
it is equivalent, wikipedia actually has this as an example. i figured you wanted to show it though. i'd use a truth table. do you know how to construct one?

5. yes I do know how, actually my question is to show

-(p+q) and p biimplication q are logically equivalent, and I am trying to find out how

6. Originally Posted by EquinoX
yes I do know how, actually my question is to show

-(p+q) and p biimplication q are logically equivalent, and I am trying to find out how
Here's how to show the first one, maybe you can figure out from this how to show the second. Truth tables are the conventional way to test whether two statements are logically equivalent. Look at the truth table below. You will notice that the blue columns are the same. One represents P + Q and the other represents (P v Q) ^ ~(P ^ Q). if two statements have the same truth values in every instant in a truth table, then they are logically equivalent

7. my teacher asks me not to use the truth table this time but to change the expression into other expression

8. Recall that $p \Leftrightarrow q\quad \equiv \quad \left( { \sim p \wedge \sim q} \right) \vee \left( {p \wedge q} \right)$.

Thus
$\begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \sim \left[ {\left( {p \vee q} \right) \wedge \sim \left( {p \wedge q} \right)} \right] \\
& \equiv & \left[ { \sim \left( {p \vee q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\end{array}$

$\begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \left[ {\left( { \sim p \wedge \sim q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\; & \equiv & p \Leftrightarrow q \\
\end{array}$
.

9. how did you change the sign of ^ into V in proofing the - (p + q)

10. Originally Posted by EquinoX
how did you change the sign of ^ into V in proofing the - (p + q)
be specific as to what line you are talking about

11. sorry it's posted below, I dont understand on how to get the ^ symbol changes into v

12. Thus
$\begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \sim \left[ {\left( {p \vee q} \right) \wedge \sim \left( {p \wedge q} \right)} \right] \\
& \equiv & \left[ { \sim \left( {p \vee q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\end{array}$

13. $
\begin{array}{l}
\sim \left( {A \wedge B} \right) \equiv \left[ { \sim A \vee \sim B} \right] \\
\sim \left( {A \vee B} \right) \equiv \left[ { \sim A \wedge \sim B} \right] \\
\end{array}
$

14. thanks that clears me up