Question about quantifiers

[Mr Davis 97]

I have the following statement: If $\phi : G \to H$ is an isomorphism, then $|\phi(x) | = |x|$ for all $x \in G$. Is this equivalent to the following? There exists an $x \in G$ such that if $\phi : G \to H$ is an isomorphism, then $|\phi(x) | = |x|$.

 

[fresh_42]

I have the following statement: If $\phi : G \to H$ is an isomorphism, then $|\phi(x) | = |x|$ for all $x \in G$. Is this equivalent to the following? There exists an $x \in G$ such that if $\phi : G \to H$ is an isomorphism, then $|\phi(x) | = |x|$.

They are both always true, so in a way they are equivalent. What do you really want to know? As soon as you have the requirement "isomorphism" it doesn't matter anymore whether there is just one $x$, because there are always all.

 

[Mr Davis 97]

They are both always true, so in a way they are equivalent. What do you really want to know? As soon as you have the requirement "isomorphism" it doesn't matter anymore whether there is just one $x$, because there are always all.

I'm just confused about how the two statement are saying the same thing. Basically what I am trying to do is to put the statement into prenex normal form and then to interpret how the statements are equivalent. I'm using the rule for taking the quantifier of the consequent that is described here: https://en.wikipedia.org/wiki/Prenex_normal_form#Implication

Edit: Actually, I used the wrong rule to begin with, so in this case my question is meaningless. If I come across another example I'll ask.

 

[ad infinitum]

The first statement says that for all x (it is true that if phi is an isomorphism then abs(phi(x))=abs(x)). The second statement says that for at least one x (it is true that if phi is an isomorphism then abs(phi(x))=abs(x)). If the first statement is true for all x then certainly it is true for at least one x (with the condition that the set of solutions for x in the formula abs(phi(x))=abs(x) must be nonempty.In the converse case the first statement only follows from second if the size of the solution set for abs(phi(x))=abs(x) is less than equal or less than one.