# 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. – user935 Mar 6 '18 at 4:54

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.