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

Here $p(x)$ and $q(x)$ are first order formulae with $x$ as their free variable

  1. $\Big( \forall x[p(x) \Rightarrow q(x)] \Big) \Rightarrow \Big(\forall x[p(x)] \Rightarrow \forall x[q(x)] \Big)$
  2. $ \Big(\forall x [p(x)] \Rightarrow \forall x[q(x)] \Big) \Rightarrow \Big(\forall x[p(x) \Rightarrow q(x)]\Big)$

My (not-so-sound)reasoning is as follows

  1. It is given that whenever $p(x)$ is true for any value of $x$ in the universe $q(x)$ is also true$\Big( \forall x[p(x) \Rightarrow q(x)] \Big)$, Thus if $p(x)$ is true for the entire universe, $q(x)$ will also be true for the entire universe and $\Big(\forall x[p(x)] \Rightarrow \forall x[q(x)] \Big)$ is true
  2. It is given that if $p(x)$ is true for the entire universe, $q(x)$ will also be true for the entire universe$\Big(\forall x[p(x)] \Rightarrow \forall x[q(x)] \Big)$. I am thinking that it need not be the case that $q(x)$ is true for cases when $p(x)$ is true.

Is my reasoning correct? I am confused and think that I am merely juggling words around. Could you give me an example to show option 2 is not valid.

share|cite|improve this question
I corrected what I took to be a typo: the antecedent of your second statement read: $\forall x[p(x)] \rightarrow \forall x [p(x)]$. I am assuming you meant for the second "p(x)" to type "q(x)". Correct me if I'm wrong. – amWhy Nov 20 '12 at 2:35
@amWhy, That's right. Thanks. – Abhijith Nov 20 '12 at 2:37
up vote 1 down vote accepted

Your reasoning in the first question is correct.

For the second question, let our universe be the set of natural numbers. Let $p(x)$ be the assertion $x$ is even, and let $q(x)$ be the assertion $x$ is a perfect square.

Then $\forall x\,p(x)$ and $\forall x \,q(x)$ are both false, and therefore the implication $\forall x \,p(x)\Rightarrow \forall x \,q(x)$ is true.

However, $\forall x\left(p(x)\Rightarrow q(x)\right)$ is false.

Many examples can be constructed along these lines, none very interesting.

share|cite|improve this answer
Thanks. I wasn't thinking about this from the false-false perspective. – Abhijith Nov 20 '12 at 3:07
Sorry about my earlier comment; I've made some silly mistakes today! – amWhy Nov 20 '12 at 3:10
@Abhijith: Now you can see why I wrote "none very interesting." – André Nicolas Nov 20 '12 at 3:12

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.