Logical equivalence problem!

[Frenchie]

Hello all, I'm seeking help in logical equivalence for the following:

 

 

{[(p^q)V(p^r)]V(q^r)}^~[(p^q)^r]

 

 

It's asking to have an equivalence of that equation using only negations of conditionals(biconditionals too I assume because of the v)

 

 

First I take (p∧q), (p∧r), and (q∧r) and (p∧q) which are all conjunctions and change them into the negation of a biconditional based on ^ being the same as <-> which is a conditional so that they are equivalent.

~(p->~q), ~(p->~r), ~(q->~r), ~(p->~q) which gives me the new equation of {[~(p->~q)v~(p->~r)v~(q->~r)]}∧~[~(p->~q)∧r]

 

I then take the remaining ∧ which are {[~(p->~q)v~(p->~r)v~(q->~r)]}∧~[~(p->~q)∧r] and ~[~(p->~q)∧r and do the same as previously.

 

I then take the disjunction two at a time starting with ~(p->~q)v~(p->~r) and change it into the negation of a conditional based on v being the same as -> which is a biconditional so that they are equivalent. I am not sure how to do this step.

 

Afterwards I would take the result of the negation of my conditional which is a biconditional as done above and do it's negation of conditional with the remaining ~(q->~r). I am not sure how to do this step.

 

I appreciate all the help that i can receive, thank you.

Thank you for any help!

 

 

Edited January 24, 2012 by Frenchie

[Frenchie]

I really need help on this one if anyone can that would be great : /

[DrRocket]

Hello all, I'm seeking help in logical equivalence for the following:

 

 

{[(p^q)V(p^r)]V(q^r)}^~[(p^q)^r]

 

 

It's asking to have an equivalence of that equation using only negations of conditionals(biconditionals too I assume because of the v)

 

 

 

 

 

Maybe if you draw a Venn diagram you can see an answer.

 

You are looking for everything that is in two of p,q, and r but not in all three.

[khaled]

This what I get when simplifying your statement:

 

= { [ (p^q) V (p^r) ] V (q^r) } ^ ~[(p^q) ^ r]

 

= { (p^q) V (p^r) V (q^r) } ^ { ~p V ~q V ~r }

 

= { (p^q) V (p^r) V (q^r) } ^ ~{ p ^ q ^ r }

 

So, it's just as DrRocket mentioned,

 

Check: Wolfram|Alpha:Solution

Edited January 29, 2012 by khaled