Q:

Question about "Some A is B" in logic

In logic we say,
"All A are B" to mean (x)(Ax ⊃ Bx)
"Some A is B" to mean (∃x)(Ax . Bx)
I can see how (x)(Ax ⊃ Bx) makes sense, By looking the table if we had A = 1, B = 0 then this statement wouldn't hold.
My question is how do we know that (∃x)(Ax . Bx) means "Some A is B"? What if I had (∃x)(Ax ⊃ Bx) instead? Can it be explained how does it all make sense using a truth table if possible?
A  B  | A . B | A ⊃ B
0  0  |   0   |   1
0  1  |   0   |   1
1  0  |   0   |   0
1  1  |   1   |   1

A:

The main difference between  (∃x)(Ax . Bx) and (∃x)(Ax ⊃ Bx) is the committment they make to the existence of As and Bs.  In fact, this is read directly off the truth table you gave.
You can interpret (∃x)(Ax . Bx) as saying that there is something that is both A and B.  So you know that there is at least one A and at least one B.  There is only a single 1 in the truth table, which is the case where A(x) gets the value 1 and B(x) gets the value 1.  The object you're committed to has both properties.
(∃x)(Ax ⊃ Bx) on the other hand says that there is at least one thing such that if it is A, then it is B.  All of the 1s in the truth table describe consistent cases. You don't necessarily know that there is something that is A; perhaps for every object, A(x) gets the value 0, so (Ax ⊃ Bx) works out true.  In fact, you only need there to be one thing such that A(x) gets the value zero to make (∃x)(Ax ⊃ Bx) come out as true.
You can read (∃x)(Ax ⊃ Bx) as equivalent to (∃x)(¬Ax v Bx), which is in turn equivalent to (∃x)(¬Ax) v (∃x)(Bx).  So the commitments we make are only to either there being some object that is not A, or there being some object that is B.
"Some A is B" has the former commitments rather than the latter, because we're saying that there is an A, such that it is also B.

