Why can't we use implication for the existential quantifier?
To me it says there is some fruit that if it is an apple, it is delicious.
This is absolutely correct. There exists a fruit such that if it is an apple, then it is delicious. Let $x$ be such a fruit. We have two cases for what $x$ may be here:
- $x$ is an apple. Then $x$ is delicious. This is the $x$ you are searching for.
- $x$ is not an apple. Now the statement "if $x$ is an apple, then $x$ is delicious" automatically holds true. Since $x$ is not an apple, the conclusion doesn't matter. The statement is vacuously true.
So the statement $\exists{x} \in F, A(x) \implies D(x)$ fails to capture precisely your desired values of $x$, i.e., apples which are delicious, because it also includes other fruits.
I'm not quite sure that I really understand WHY I need to use implication for universal quantification, and conjunction for existential quantification.
Modifying your analysis a bit, let $A$ be the set of apples, and $D$ the set of delicious things.
$\forall x: [x\in A \implies x\in D]$ means all apples are delicious. Often written $\forall x\in A :x\in D$
$\forall x: [x\in A \land x\in D]$ means everything is a delicious apple.
$\exists x:[x\in A \land x\in D]$ means there exists at least one delicious apple. Often written $\exists x\in A:x\in D$, or equivalently $\exists x\in D: x\in A$
What does $\exists x:[x\in A \implies x\in D]$ mean? It is equivalent to $\exists x:[x\notin A \lor x\in D]$.
For a given $x$ then, either of the following possibilities that will satisfy this condition:
$x\in A \land x\in D$, i.e. there exists at least one delicious apple (as above)
$x\notin A\land x\in D$, i.e. there exists at least one non-apple that is delicious
$x\notin A\land x\notin D$, i.e. there exists at least one non-apple that is not delicious
So, the implication allows for more possibilities than the conjunction. In particular, the implication allows for the possibility that there are no apples. The conjunction does not.
Furthermore, $\exists x: [x\in A \implies x\in D]$ is a set theoretic variation of the so-called Drinker's Paradox. Here's where it gets crazy! For any set $A$ and any proposition $P$, we can prove using ordinary set theory that $$\exists x: [x\in A \implies P]$$
You could even prove, for example, that $$\exists x: [x\in A \implies x\notin A]$$ So, to avoid confusion, you would probably want to avoid such constructs in mathematics. For a formal development, see The Drinker's Paradox: A Tale of Three Paradoxes at my blog.
$\forall x\in F (A(x)\to D(x))$ versus $\exists x\in F(A(x)\wedge D(x))$
$\forall x\in F (A(x)\to D(x))$ says "any fruit, if it is an apple, then it is delicious," or simply, "apples are delicious fruit".
$\forall x\in F (A(x)\wedge D(x))$ says "any fruit, is an apple and is delicious", or simply "all fruit are delicious apples."
Which says what you mean?
$\exists x\in F(A(x)\wedge D(x))$ says "some fruits, is an apple and is delicious," or simply "there is a delicious apple".
$\exists x\in F (A(x)\to D(x))$ says "there is some fruit, that if it were an apple then it would be delicious". Witness this mouldy orange.
Which says what you mean?