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

Printable View

- Jun 11th 2007, 08:42 AMEquinoXis this logical expression equivalent?
p + q = (pVq) ^ -(p^q)

- Jun 11th 2007, 08:47 AMJhevon
- Jun 11th 2007, 08:50 AMEquinoX
it means an exclusive or

- Jun 11th 2007, 08:53 AMJhevon
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, 08:59 AMEquinoX
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, 09:03 AMJhevon
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, 09:08 AMEquinoX
my teacher asks me not to use the truth table this time but to change the expression into other expression

- Jun 11th 2007, 09:26 AMPlato
Recall that $\displaystyle p \Leftrightarrow q\quad \equiv \quad \left( { \sim p \wedge \sim q} \right) \vee \left( {p \wedge q} \right)$.

Thus

$\displaystyle \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}$

$\displaystyle \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, 09:43 AMEquinoX
how did you change the sign of ^ into V in proofing the - (p + q)

- Jun 11th 2007, 09:49 AMJhevon
- Jun 11th 2007, 09:52 AMEquinoX
sorry it's posted below, I dont understand on how to get the ^ symbol changes into v

- Jun 11th 2007, 09:54 AMEquinoX
Thus

$\displaystyle \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, 10:09 AMPlato
$\displaystyle

\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, 10:12 AMEquinoX
thanks that clears me up