Here $p(x)$ and $q(x)$ are first order formulae with $x$ as their free variable
- $\Big( \forall x[p(x) \Rightarrow q(x)] \Big) \Rightarrow \Big(\forall x[p(x)] \Rightarrow \forall x[q(x)] \Big)$
- $ \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
- 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
- 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.