# Which of the following is a valid first order formula?

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.

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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

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.

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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