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
• Your terminology "how does logical proposition X hold to be true" is quite difficult to interpret. Is there a particular proof, deduction or situation you're trying to evaluate? Could you explain what it is about those situations that you're not quite following? – Paul Ross Nov 1 '12 at 16:28
• @PaulRoss: I edited the question. I can understand how (x)(Ax ⊃ Bx) comes to mean "All A are B". I'm confused with how (A . B) is false in 3 cases and (A ⊃ B) has only one case that is false. So, I'm trying to understand how can (∃x)(Ax . Bx) make sense if I look at the cases in table if you could explain that. – user963241 Nov 1 '12 at 16:35
• The problem is that a truth table can only reveal what is true of a single object x. In order to tell whether there even exists an A, a truth-table doesn't reveal anything: whereas you can evaluate (∀x)(Ax ⊃ Bx) by simply restricting to the rows of the table where A holds, and see whether B also holds, the only thing a truth table reveals about (∃x)(Ax . Bx) is that there are some meanings of A and B in which it is possible for some A to also be B. Rather than a truth table, you need an table of objects, to see whether each satisfied A and/or B. – Niel de Beaudrap Nov 1 '12 at 20:47
• I disagree with your translation of "some A are B", because it's theoretically possible A(x) never holds at all. My translation: if there exists x such that A(x) THEN there exists y (which may or may not be equal to x) such at A(y) implies B(y). Your version assumes there exists at least one x such that A(x) holds. – barrycarter Mar 6 '18 at 4:54