# is this logical expression equivalent?

• Jun 11th 2007, 09:42 AM
EquinoX
is this logical expression equivalent?
p + q = (pVq) ^ -(p^q)
• Jun 11th 2007, 09:47 AM
Jhevon
Quote:

Originally Posted by EquinoX
p + q = (pVq) ^ -(p^q)

what does the + mean? have you tried doing a truth table?
• Jun 11th 2007, 09:50 AM
EquinoX
it means an exclusive or
• Jun 11th 2007, 09:53 AM
Jhevon
Quote:

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?
• Jun 11th 2007, 09:59 AM
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
• Jun 11th 2007, 10:03 AM
Jhevon
Quote:

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
• Jun 11th 2007, 10:08 AM
EquinoX
my teacher asks me not to use the truth table this time but to change the expression into other expression
• Jun 11th 2007, 10:26 AM
Plato
Recall that $p \Leftrightarrow q\quad \equiv \quad \left( { \sim p \wedge \sim q} \right) \vee \left( {p \wedge q} \right)$.

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

$\begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \left[ {\left( { \sim p \wedge \sim q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\; & \equiv & p \Leftrightarrow q \\
\end{array}$
.
• Jun 11th 2007, 10:43 AM
EquinoX
how did you change the sign of ^ into V in proofing the - (p + q)
• Jun 11th 2007, 10:49 AM
Jhevon
Quote:

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
• Jun 11th 2007, 10:52 AM
EquinoX
sorry it's posted below, I dont understand on how to get the ^ symbol changes into v
• Jun 11th 2007, 10:54 AM
EquinoX
Thus
$\begin{array}{rcl}
\sim \left( {p + q} \right) & \equiv & \sim \left[ {\left( {p \vee q} \right) \wedge \sim \left( {p \wedge q} \right)} \right] \\
& \equiv & \left[ { \sim \left( {p \vee q} \right) \vee \left( {p \wedge q} \right)} \right] \\
\end{array}$
• Jun 11th 2007, 11:09 AM
Plato
$
\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}
$
• Jun 11th 2007, 11:12 AM
EquinoX
thanks that clears me up