I was doing an exercise that said convert the statement "Jane saw a police officer, and Rodger saw one too" into the logical equivalent using quantifiers. My answer was: $$ \exists x(P(x)\implies S(j,x)) \land \exists y(P(y)\implies S(r,y)) $$ where $P(x)$ stands for "$x$ is a police officer", $S(x,y)$ stands for "$x$ saw $y$" with $j$ & $r$ standing for Jane and Rodger.
But the answer in the back of the book is:
$$ \exists x(P(x)\land S(j,x)) \land \exists y(P(y)\land S(r,y)) $$
Obviously the statements aren't logically equivalent but it seems that the statement $P(x)\implies S(j,x)$ is stating that "if $x$ is police officer then Jane saw $x$" which is the same as $P(x)\land S(j,x)$ being true. I'm guessing the problem is that the truth table values for the implication $A\implies B$ in the cases where it's not false (False implies True and False implies False) and thus differ from the truth table for $A \land B$ but is that relevant to this application since to me it seems we are only interested in the cases that make $P(x)$ (and $S(x,y)$) true and thus eliminating the cases that make the truth tables for "$\implies$" and "$\land$" differ. But I suppose that since we are stating that there exists an $x$ such that $P(x)\implies S(j,x)$ couldn't that mean the $x$ may be the case that makes both $P(x)$ and $S(j,x)$ false but consequently making the implication true.
So I guess my question is why aren't these two statements equivalent in this application? I'm still only just beginning logic so I've probably overlooked something or I don't fully understand it.