My Math Forum Simplifying Algebra Proposition

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 October 9th, 2011, 10:19 AM #1 Newbie   Joined: Oct 2011 Posts: 4 Thanks: 0 Simplifying Algebra Proposition guys i seriously have no idea on how to start this one.. (pVq)^(-pVq)^(pV-q)^(-pV-q) how to simplify this one? my teacher gave me this example: p^(pVq) =- (pVf)^(pVq) <--- how is that f in there?? =- p V (f^q) =- p V f =- p i could just solve this using truth table, but he want me to use algebra rules. if you guys can help me, i would definitely appreciate it thanks alot im having problem with the rules of algebra being implemented in non numeric arguments. i did truth table of (pVq)^(-pVq)^(pV-q)^(-pV-q) and found that the whole proposition is equivalent to (-pV-q)
October 9th, 2011, 04:19 PM   #2
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Re: Simplifying Algebra Proposition

Quote:
 Originally Posted by abang526 my teacher gave me this example: p^(pVq) =- (pVf)^(pVq) <--- how is that f in there?? =- p V (f^q) =- p V f =- p
There is a negation missing in the first line.

Your teacher uses $F$ to denote the truth value of a false statement, which he meant a contradiction since a contradicting statement is false.

I assume you know that the truth value of $T\vee F \equiv T\quad ^{[*]}$, and of $T\wedge F\equiv F\qua ^{[**]}$.

If you substitute $P$ for $T$, you will get $P\vee F \equiv P$, and of $P\wedge F\equiv F$

$\sim P \wedge (P\vee Q)\equiv \sim (P\vee F)\wedge (P\vee Q)$, see[*]

$F\wedg Q \equiv F$, see [**]
Quote:
 Originally Posted by abang526 guys i seriously have no idea on how to start this one.. (pVq)^(-pVq)^(pV-q)^(-pV-q) how to simplify this one?
You can rearrange it so that you can get similar terms for contraction. Like this $\left [(P\vee Q)\wedge (P\vee \sim Q)\right ] \wedge \left [(\sim P\vee Q) \wedge (\sim P\vee \sim Q)\right ]$

You do the rest.

 October 9th, 2011, 04:59 PM #3 Newbie   Joined: Oct 2011 Posts: 4 Thanks: 0 Re: Simplifying Algebra Proposition [ ((p ^ -p) V q) ^ (p V -q) ^ (-p V -q) ((p ^ -p) V q) ^ ((p ^ -p) V -q) (F V q) ^ ((p ^ -p) V-q) = (F V q = q) ^ ( (p ^ -p = F) V -q) = q ^ (F V -q) = q ^-q did i do it right?
October 9th, 2011, 07:45 PM   #4
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Re: Simplifying Algebra Proposition

Hello, abang526!

The statement is a self-contradiction . . . always false.

We need a couple of theorems:

[color=beige]. . [/color]$\begin{array}{cccccc}\text{Theorem 1:}=&s\,\wedge\,\sim s=&F \\ \\ \\ \text{Theorem 2:}=&F\,\vee\,s=&s \end{array}=$

Quote:
 $(p\,\vee\,q)\:\wedge\:(\sim p\,\vee\,q)\:\wedge\:(p\,\vee\,\sim q) \:\wedge\:(\sim p\,\vee\,\sim q)$

$\begin{array}{cccccccc} 1. & (p\,\vee\,q)\:\wedge\:(\sim p\,\vee\,q)\:\wedge\:(p\,\vee\,\sim q) \:\wedge\:(\sim p\,\vee\,\sim q) &&& 1. & \text{Given} \\ \\ \\ \\ 2. & [(p\,\wedge\,\sim p)\,\vee\,q] \:\wedge\:[(p\,\wedge\,\sim p) \,\vee\,\sim q] &&& 2. & \text{Distr.} \\ \\ \\ \\ 3. & [F\,\vee\,q]\:\wedge\:[F\,\vee\,\sim q] &&& 3. & \text{Theorem 1} \\ \\ \\ \\ 4. & q\:\wedge\:\sim q &&& 4. & \text{Theorem 2} \\ \\ \\ \\ 5. & F &&& 5. & \text{Theorem 1} \end{array}$

 October 9th, 2011, 10:32 PM #5 Newbie   Joined: Oct 2011 Posts: 4 Thanks: 0 Re: Simplifying Algebra Proposition thanks soroban for your explanation.. so anyway, here's something i tried on my own.. (p^-(pV-q)) V q ^ (qVp) p ^ -p ^ q V q <- ( q ^ q V p = q right? ) f ^ q V q f V q T is it correct? thanks
 October 10th, 2011, 01:52 AM #6 Newbie   Joined: Oct 2011 Posts: 4 Thanks: 0 Re: Simplifying Algebra Proposition i dont mean to keep asking here, i do get the ideas but the methods are somehow different.. im totally lost on how to do the next one.. (p V ( -q -> p)) V ((p <-> -q) -> (q ^ -p)) (p ^ (q -> -r)) V ((-p V r) <-> -q) (p ^ q) -> ( p ^ -r) V ((-p V -q) <-> ( r V -q)) <- does it work like this? thanks a lot!!
 September 29th, 2013, 03:42 AM #7 Newbie   Joined: Sep 2013 Posts: 1 Thanks: 0 Re: Simplifying Algebra Proposition how to simplyfy this? (p'^q)\/('(p^q')^(p\/q'))

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