Using Implication vs Conjunction in predicate logic Consider the following question.
In the domain of all people, let F(x) be “x is your friend,” and let
P(x) be “x is perfect.”
Translate each of these statements into logical expressions using predicates, quantifiers, and
logical connectives.
a) All your friends are perfect.
b) At least one of your friends is perfect.
I tried this and came up with the following solutions.
a) ∀x(F(x) → P(x))
b) ∃x(F(x) → P(x))
The answer of a) is same as given in the key. However, for b), the answer given is ∃x(F(x) ∧ P(x)).
I am not able to understand why conjunction is being used instead of implication in the second case. Also is there any rule as to when implication/conjunction must be used?
 A: 
Why your (b) is wrong?

Consider the case where there are no friends: this means that $Fx$ is False, and thus the conditional $Fx \to Px$ is True.
Thus, if there are no friends, $\exists x (Fx \to Px)$ will be True, contrary to expectations.
The formula $\exists x (Fx \land Px)$ instead, will be False when there are no friends, and this is consistent with the intuitive reading of "At least one of your friends is perfect".
A: The truth table for $p\rightarrow q$, is:
$$\begin{array}{|c|c|c|}
\hline
p & q & p\rightarrow q \\ \hline
true  & true  & true \\ \hline
true  & false & false\\ \hline
false & true  & true \\ \hline
false & false & true \\ \hline
\end{array}$$
So, whenever $p$ is false, $p\rightarrow q$ is true. Let $\alpha$ be not your friend. That is: $\neg F(\alpha)$. Then, $F(\alpha)$ is false and $F(\alpha)\rightarrow P(\alpha)$ is true. By existential generalization, we get: $\exists x\left(F(x)\rightarrow P(x)\right)$. This is true, even if none of your friends is perfect.
On the other side, the truth table for $p\wedge q$, is:
$$\begin{array}{|c|c|c|}
\hline
p & q & p\wedge q \\ \hline
true  & true  & true \\ \hline
true  & false & false\\ \hline
false & true  & false \\ \hline
false & false & false \\ \hline
\end{array}$$
So, $F(\alpha)\wedge P(\alpha)$ will be true, only if both $F(\alpha)$ and  $P(\alpha)$ are true, that is, only if $\alpha$ is both your friend and perfect. Then, $\exists x\left(F(x)\wedge P(x)\right)$ means that there exist at least one thing that is your friend and it is perfect.
