(p <--> q) and not q - Question

**madmunk** · June 28th 2008, 07:47 AM

Hello,

I just have a quick question about the statement in the title.

(p <-> q) AND NOT q

If I was to find the truth statement from that.

How am I to break the (p <-> q) down into a (p AND q) type statement first?

because I assume that if I can get the q out of there, then q AND NOT q together will = False.

And then any AND statement with a false in it is inevitably False.

Thanks

**Moo** · June 28th 2008, 08:04 AM

Hello,

$\wedge=and$
$\vee=or$
$\bar{a}=non \ a$

$p \leftrightarrow q$ is $\{p \rightarrow q\} \wedge \{q \rightarrow p\}$

$\{p \rightarrow q\}$ is $\bar{p} \vee q$
$\{q \rightarrow p\}$ is $\bar{q} \vee p$

So $p \leftrightarrow q$ is $\{\bar{p} \vee q\} \wedge \{\bar{q} \vee p\}$

If you want the formula of your proposition, develop, according to these rules :

$(a \wedge b) \vee c=(a \vee c) \wedge (b \vee c)$

and $(a \vee b) \wedge c=(a \wedge c) \vee (b \wedge c)$

and the associativity of $\vee$ and $\wedge$

------------------

Your reasoning with "necessarily false" is not correct, because it can be "or" instead of "and".

**madmunk** · June 28th 2008, 08:16 AM

Thanks Moo, I see what I missed now ! I was looking at the answer the whole time. I should have been able to figure out that to break it up it looked like that !

Thanks for the reasoning part as well

I think I might have to hit the books again and get a bit more practise under the belt !

Cheers

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