You may have noticed that in the exercises in 10.1, sometimes there was more than one way to express statements in the notation of predicate logic. For instance, just as in English we could express the statement “Not all jazz fans like Thelonious Monk” by saying “Some jazz fans don’t like Thelonious Monk,” in predicate logic, we can say that

~(x) (Jx > Lx) says the same thing as (∃x) (Jx . ~Lx).

A little thought makes this clear, since the first is the negation of an A statement, while the second is an O statement. Since A and O are contradictories of one another, whenever one is true the other is false: either one is the same as the negation of its contradiction.

This is a very useful thing to know, and not only for doing proofs in this class. It is also useful outside this class, in helping you have control over understanding what you or others are saying.

But for the purposes of this study of predicate logic, it comes down to the principle that any statement *can be rewritten as* (is equivalent to) the negation of its contradiction. Why this is useful is because the main operators of the two versions are not the same: in the one version, the main operator is a tilde; in the other, it is a quantifier. We have rules that guide us in removing and adding quantifiers. So knowing that every tilde-governed statement is equivalent to some quantifer-governed statement means a new dimension of flexibility.

“All Popes are Catholics” is equivalent to “It is false that Some Popes are not Catholics”

“No Jews are Hindus” is equivalent to “It is false that *any* (which is the form that “some” takes in English when used in negative contexts) Jews are Hindus.”

“Some Philosophers are Existentialists” is equivalent to “It is false that no Philosophers are Existentialists.”

“Some Civil Rights leaders are not men” is equivalent to “It is false that all Civil Rights leaders are men.”

Let’s represent these equivalence examples in our notation:

(x) (Px > Cx) :: ~(∃x) (Px . ~Cx)

(x) (Jx > ~Hx) :: ~(∃x) (Jx . Hx)

(∃x) (Px . Ex) :: ~(x) (Px > ~Ex)

(∃x) (Cx . ~Mx) :: ~(x) (Cx > Mx)

What happens in these equivalences can be summed up as two moves :

a) you change the quantifier of a statement (from universal to particular or vice versa), and

b) you add a tilde **in front of and behind** the quantifier.

Schematically:

(x) Fx :: ~(∃x) ~Fx “All are” is the same as “it’s false that any are not.” (The left hand expression here is the standard “A.”)

~(x) Fx :: (∃x) ~Fx “Not all are” is the same as “Some are not.” (The right hand here is “O.”)

(∃x) Fx :: ~(x) ~Fx “Some are” is the same as “it’s false that none are.” (The left hand here is “I.”)

~(∃x) Fx :: (x) ~Fx “Not any are” (or “it’s false that any are”) is the same as “None are ” (The right hand here is “E.”)

Any of these four equivalences (working from either side to the other side) is justified by the rule called **CQ** : Change of Quantifier

One remarkable little fact this exhibits is the ambiguity of the word “any.” In some contexts “any” means “all,” as when you say “anyone who votes is a citizen” (“all people who vote are citizens”). In other contexts, “any” means “some,” as when you say “if any one asks you where I am, don’t tell him.” This clearly means “if *some* person asks you.”

We’ve made much of the difference between “ALL” and “SOME” throughout this course. Here we become aware that our ordinary language lets the word “ANY” have both of these meanings, which are completely incompatible, in different sorts of contexts!

*Anytime you hear anyone use the word “any,” you have to wonder if she is trying to say “all” or “some.”*

In the preceding statement, the first use of “any” (in “anytime”) meant “all,” but the second one (in “anyone”) meant “some.”

*For any x, if x is a time, then if there is any y, such that y says “any,” then for any z, if z is another person, z has to wonder what y meant.*

(x) Tx > [ (∃y) Sy > (z) (Pz > Wzy)]

As the predicate notation makes clear, the first occurrence of “any” in this statement is universal, the second is particular, and the fourth is universal again ! What about the third? The third occurrence of “any” is not a case of *use*, it is a case of *mention*. So it doesn’t get represented here: it gets “swallowed up” in “Sy,” which means “y says ‘any’.” To review the *use/mention *distinction, go back to the first chapter. (It’s time to do that anyway, the exam is next week!)

An Observation:

You might recall that when we first saw how to write A and E statements in predicate notation, and I and O statements, I insisted that the main operator inside the parentheses or brackets for A or E will always be a horseshoe. Likewise, I insisted that the main operator inside the parentheses or brackets of I and O statements will be a dot. Now, when you apply the two-part rule of CQ, it might appear that this generalization is being abandoned. Take a look:

Let’s rewrite (x) (Mx > Ox) by the CQ rule. 1. Change the quantifier, 2. Put a tilde before and after it.

~(∃x) ~(Mx > Ox)

Now doesn’t that look like an existential statement with a horseshoe instead of a dot as its main operator? I hope you can see that it isn’t, since it’s a *negated* existential statement (the main operator is not the quantifier but the tilde.)

But even so, let’s see what happens as we apply some rules of equivalence to the part of the expression coming after the quantifier:

~(∃x) ~( ~Mx v Ox) this is Material Implication, changing the horseshoe to a wedge.

~(∃x) (Mx . ~Ox) and this is DM and DN, changing the negated disjunction to a conjunction of negatives (and removing the double negation).

So it turns out that what appeared to be a horseshoe statement form in an existential statement is not distinct from a dot after all. They are equivalent, and we can see why, not just take it on someone’s authority.

Some Exercises:

You will need to use the rule called CQ (Change of Quantifier) to do these proofs. This is a rule of equivalence, so you can rewrite a quantified statement wherever it occurs (it need not be the entire line that gets rewritten). And, you can just say “CQ” for any of the above four equivalences.

Exercises

1)

1. (x) Ax > (∃x) Bx

2. (x) ~Bx / (∃x) ~Ax

2)

1. (∃x) ~Ax v (∃x) ~Bx

2. (x) Bx / ~(x) Ax

3)

1. ~(∃x) Bx / (x) (Bx > Cx)

4)

1. (x) Ax > (∃x) ~Bx

2. ~(x) Bx > (∃x) ~Cx / (x) Cx > (∃x) ~Ax

5) If all philosophers are either ethicists or metaphysicians, then there are no logicians. But Russell’s a logician, so some philosophers are not metaphysicians.

6) Either Leibniz is a phenomenologist or all the existentialists are phenomenologists. But it is not the case that there are any phenomenologists, therefore Leibniz is not an existentialist.

7) All utilitarians are ethicists and all idealists are metaphysicians. Therefore, since it is not true that some ethicists are metaphysicians, it is not the case that some utilitarians are idealists.

8) All poorly trained logicians and ethicists are untrustworthy philosophers. But it is not the case that any philosophers are untrustworthy, so it is not the case that any ethicists are poorly trained.