is this logical expression equivalent?

**EquinoX** · June 11th 2007, 08:42 AM

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

**Jhevon** · June 11th 2007, 08:47 AM

Originally Posted by EquinoX

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

what does the + mean? have you tried doing a truth table?

**EquinoX** · June 11th 2007, 08:50 AM

it means an exclusive or

**Jhevon** · June 11th 2007, 08:53 AM

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?

**EquinoX** · June 11th 2007, 08:59 AM

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

**Jhevon** · June 11th 2007, 09:03 AM

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

**EquinoX** · June 11th 2007, 09:08 AM

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

**Plato** · June 11th 2007, 09:26 AM

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

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

**EquinoX** · June 11th 2007, 09:43 AM

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

**Jhevon** · June 11th 2007, 09:49 AM

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

**EquinoX** · June 11th 2007, 09:52 AM

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

**EquinoX** · June 11th 2007, 09:54 AM

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

**Plato** · June 11th 2007, 10:09 AM

$ \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} $

**EquinoX** · June 11th 2007, 10:12 AM

thanks that clears me up

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