Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|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
How about a truth table? That would be the simplest thing to do, I think. – Arturo Magidin Sep 23 '10 at 19:58
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
@Noldorin: Yes, they are not equivalent. See my answer. – Aryabhata Sep 24 '10 at 1:20
It seems clear to me that it should have been $(p\rightarrow q)\wedge (p\rightarrow r)$ on the right hand side. I checked to make sure it wasn't me who introduced the mistake when I edited the question, and it was incorrect in the original. – Arturo Magidin Sep 24 '10 at 13:53

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|cite|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|cite|improve this answer

Your Answer


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.