Against both my better judgment and the hue and cry of many, I will continue my semi-informed-by-past-years-of-studying "exposition" of predicate logic which I started here. If I accomplish nothing else, I will give Burl something to complain about for the next week or so.
In the previous installment, we talked about how syllogistic statements about "all x's" assert the truth of a conditional statement. "All dogs bark" asserts that for all x's, if x is a dog, then x barks. Formally expressed, that's:
∀x(Dog(x) –> Barks(x))
or something similar. It doesn't say anything about whether there actually are any dogs. Additionally, the 'For all...' symbol - ∀ - doesn't allow you to say anything about only some dogs. Let us address that issue.
The ∀ symbol is called a quantifier. In fact, it's called the universal quantifier. Can you guess why? Read the Wikipedia entry on quantification if you'd like more detail. There is another quantifier for when you want to say 'some', which is called the existential quantifier. Its symbol looks like this - ∃. Let's see it in action.
If we take the statement above "All dogs bark" and instead assert that "Some dogs bark," we have to say:
∃x(Dog(x) AND Barks(x))
You don't use the "if-then" arrow here, because you're not saying "There is something and, if it's a dog, then it barks." That sounds like it means the same as "all dogs bark." Instead of "if-then," we use "AND". Now, this statement is compatible with there being only one dog that barks, or all of the dogs barking, or just some of them. It doesn't tell you, for any given dog you meet, whether it barks or not.
Just as you can assert that a proposition is true and its opposite is false, you can also restate the false proposition negatively to make it true. For example, if you say that 'All dogs bark' is false, you can also restate that as 'Not all dogs bark' and say that it is true. This type of negation can be captured using symbolism as well (I'll use the tilde "~" to represent "not"):
∀x(Dog(x) –> Barks(x)) is false
~∀x(Dog(x) –> Barks(x)) is true
So when you say 'not all dogs bark', you are kind of saying that while some might, some definitely don't. It turns out negating a ∀ (universal) statement is the equivalent of asserting a related ∃ (existential) statement. "Not all dogs bark" is the same as "Some dogs don't bark," or:
~∀x(Dog(x) –> Barks(x)) ...is equivalent to ∃x(Dog(x) AND ~Barks(x))
To reiterate: the universal statement doesn't actually state the existence of anything, because it's a conditional: if something is a dog, it will bark. The negated universal statement which is an existential statement does (implicitly) assert the existence of something.
This is major difference between the two statements that we'll explore further in a future post.
--Initiated by Seth & rescued by Mark from Seth's confusion and drunkenness