Problem translating predicate syllogistic I have a question about the following predicates:
No A are B
$$
\neg \exists x (Ax\wedge Bx)
$$
For the above, my question is: why can't I write the following?
$$
\neg \exists x (Ax\rightarrow Bx)
$$
How would this one be translated to English?

Second predicate:
Not all A are B
$$
\neg \forall x (Ax\rightarrow Bx)
$$
Why can't I write:
$$
\neg \forall x (Ax\wedge Bx)
$$
Is this because in this example, the first one does say that, because some B can be A, you can't put the conjunction sign?
 A: For the first one. Suppose that $\forall x (\neg Ax)$. Then the first sentence in true and the second is false. They are just different sentence and the first one does not imply the second one. The second is similar.
Maybe is usefull to note that $A\to B$ is the same as $\neg A\vee B$. And also that $\neg \exists x (Px)$ is the same as $\forall x(\neg Px)$,  and that $\neg\forall x Px$ is the same as $\exists x(\neg Px)$. (and also $\neg(A\vee B)=\neg A \wedge\neg B$ and $\neg(A\wedge B)=\neg A\vee\neg B$) 
A: May be useful to exploit the "isomorphism" between syllogism and algebra of sets.
We have that "no $A$ is $B$" is :

$\forall x \lnot (A(x) \land B(x))$

If we introduce two sets $A$ and $B$ such that $x \in A$ iff $A(x)$ and $x \in B$ iff $B(x)$, we have that the above condition amount to :

for all $x$, not ($x \in A$ and $x \in B$),

i.e.

$A \cap B = \emptyset$.

With $\forall x \lnot (A(x) \rightarrow B(x))$ instead, we will have :

for all $x$, not (if $x \in A$, then  $x \in B$).

But in the "algebra of sets" language, $x \in A \rightarrow x \in B$, is the definition of $A \subseteq B$.
So, our formula is equivalent to :

$A \nsubseteq B$.

Now we can easy understand why the two above "translations" of "no $A$ is $B$" are not equivalent : saying that $A \nsubseteq B$ is not equivalent to $A \cap B = \emptyset$, because we may have some object $a \in A$ such that $a \notin B$ and still have some other object $b$ which is in $A$ and in $B$, so that $A \cap B \ne \emptyset$.
