Why does the form "some p(x), q(x)" expressed logically as $\exists x(A(x) \wedge P(x))$ I have searched the internet for this, but never found an explanation that I find satisfactory, in the sense that I could actually comprehend it (if there was even an explanation). There was not much explanation of this to began with.
For the following problem, where we are asked to logically express:


*

*All apples have spots.


Where the domain is all fruits, then we can create two predicates (open sentences) as follows:


*

*A(x) = "x is an apple"

*P(x) = "x has spots"


And hence the above can be expressed as: $$\forall x(A(x)\implies P(x))$$
At least, according to my distinguished professor. However if we were asked to logically express:


*

*Some apples have spots.


And using the same two predicates we defined above, P(x) and A(x), the following statement for some apples have spots: $$\exists x(A(x)\implies P(x))$$
is expressed incorrectly. Instead, the correct way to express that "some apples have spots", where the domain is fruits, is: $$\exists x(A(x) \wedge P(x))$$
So the question is, why is the above incorrect and the bottom is correct for "some apples have spots"?
Thank you.
 A: Remember that $A \to B$ can be true even if $A$ is false! If there is a spotted elephant e, then $A(e) \to P(e)$ is true, and thus $\exists x.A(x) \to P(x)$ is true too. But there might not be any spotted apples! We've only shown a spotted elephant, but somehow this proves the claim?? Clearly something has gone wrong. 
What we really want is to show that something exists which is both spotted and an apple. That is, $\exists x.A(x) \land P(x)$

I hope this helps ^_^
A: First, recall that $A\Rightarrow B$ is a shorthand for $\lnot A \lor B$.
Now, let $Q(x)$ means "$x$ is spotless", i.e. $Q(x)\Leftrightarrow\lnot P(x)$.
Your statement "some apples have spots" is obviously the negation of "all apples are spotless", i.e. it is equivalent with:
$$\begin{array}{rll}\lnot(\forall x)(A(x)\Rightarrow Q(x))&\Leftrightarrow(\exists x)\lnot(\lnot A(x)\lor Q(x))\\&\Leftrightarrow(\exists x)((\lnot\lnot A(x))\land \lnot Q(x))&\text{ DeMorgan's law}\\&\Leftrightarrow(\exists x)(A(x)\land P(x))\end{array}$$
which is the expression you were unsure about. The conjunction operation ($\land$) comes as a result of applying DeMorgan's law to the disjunction operation ($\lor$) that is implicit in the "implication" ($\Rightarrow$) operation.
A: This is because an implication is also true if the first term is false. 
So basically if $x$ is not an apple $A(x)$ is false and hence $A(x)\Rightarrow P(x)$ is true. That means, there exists an $x$ (basically all the fruits that are not apples), so that $A(x)\Rightarrow P(x)$ is true. 
But that does not necessarily mean, that there is an apple, which has spots.
A: "All apples have spots," is "Any thing, if it is an apple then it will have spots."
$$\forall x~(A(x)\to S(x))$$
"Some apples have spots," is "There is a thing that is an apple and it will have spots."
$$\exists x~(A(x)\wedge S(x))$$

$\exists x~(A(x)\to S(x))$ will claim "There is a thing, if it is an apple then it will have spots."   Witness a tiger!   It is true that "If a tiger is an apple, then it will have spots," since a conditional is only considered false when the consequent is false while the antecedent is true, and this is not the case.
So $\exists x~(A(x)\to S(x))$  is equivalent to $\exists x~(\neg A(x)\vee S(x))$ and so will be true if we have spotted things or non-apples, but have no spotted apples.   But we wish to claim there are spotted apples, so this won't do what we want it to do.
