Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the standard logical equivalence: $(p\rightarrow q)\wedge(q\rightarrow r)\Leftrightarrow p\rightarrow (q\wedge r)$.

Using several distributive laws I was able to get it down to: $(\neg p\wedge\neg q) \vee (\neg p\wedge r) \vee (q\wedge r)$.

I must be missing some manipulation I can do to reduce this.

share|improve this question
    
Where did you start, where were you trying to get? "Get it down"... get what down? –  Arturo Magidin Sep 23 '10 at 19:49
    
Well first I got rid of the conditionals by the definition of implication and then I was left with the two compound disjunctive statements with the "and" separating them. I then used distributive law twice and got to the point I stated above. What I meant by get down was that I was trying to get the compound proposition into a form so that I could conclude it was equivalent to the right side. –  Planeman Sep 23 '10 at 19:52
2  
How about a truth table? That would be the simplest thing to do, I think. –  Arturo Magidin Sep 23 '10 at 19:58
2  
I haven't worked through the details, but it strikes me that this statement is not correct. Change the bidirectional implication to a simple implication, and it holds, however. –  Noldorin Sep 24 '10 at 0:18
2  
@Noldorin: Yes, they are not equivalent. See my answer. –  Aryabhata Sep 24 '10 at 1:20

2 Answers 2

Umm... maybe I am missing something, but

if $p$ is false, $q$ is true and $r$ is false, then we have that

$(p\rightarrow q)\wedge(q\rightarrow r)$ is false

$p\rightarrow (q\wedge r)$ is true.

So I don't see how you can prove the equivalence.

share|improve this answer

$$\begin{array}{|c|c|c|c|c|c|c|c|c}\hline p&q&r&p\Rightarrow q&q\Rightarrow r&(p\Rightarrow q)\land (q\Rightarrow r)&q\land r&p\Rightarrow(q\land r)&\cdots\\\hline T&T&T&T&T&T&T&T\\T&T&F&T&F&F&F&F\\T&F&T&F&T&F&F&F\\T&F&F&F&T&F&F&F\\F&T&T&T&T&T&T&T\\F&T&F&T&F&F&F&T\\F&F&T&T&T&T&F&T\\F&F&F&T&T&T&F&T\\\hline \end{array}$$

$$ \begin{array}{c|c|c|}\hline \cdots&\big[(p\Rightarrow q)\land (q\Rightarrow r)\big]\iff \big[ p\Rightarrow(q\land r)\big]\\\hline &T\\&T\\&T\\&T\\&F\\&T\\&T\\\hline \end{array}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.