[8.] Relational Predicate Logic

 

[8.1.] "Relational Predicates." (Ch.10:1)

 

 

[8.1.1.] Relational Properties and Property Constants.

 

All of the property constants we have used up to now have represented properties possessed by single individuals:

 

“Michael is tall”                      =          Tm

“Dumbo is an elephant”                      =          Ed

“Someone is conceited”                       =          ($x)Cx

“All cows are mammals”         =          (x)(Cx É Mx)*

 

*Even in this example, the property being a cow and the property being a mammal are properties possessed by individual entities. Although there are many cows and many mammals, the property of being a cow and the property of being a mammal are possessed by one individual thing at a time.

 

But not all properties are like this. Some properties hold, not for a single individual, but between two individuals. For example:

 

·         being taller than

 

No single entity, considered all by itself, can possess the property of being taller than. It makes no sense to say that Andrew is taller than (full stop). For there to be an instance of the property being taller than (i.e., for the property being taller than to be instantiated), there must be two individual things, one of which is taller than another.

 

For example, it makes perfect sense to say that Andrew is taller than Bill. There are two individuals, Andrew and Bill, and together they instantiate the property being taller than. Other examples of two-place properties (properties that are instantiated by two individuals) are:

 

·         being shorter than

·         being the mother of (and: being the child of, or father of, or brother of)

·         being louder than (and: being quieter than)

·         being on top off (and: being under, or being beside, or being behind)

 

And some properties hold among three or more individuals. These are called three-place properties. For example:

 

·         being between (A is between B and C);

·         being given (A was given to B by C).

 

These sorts of properties, which can only be instantiated by two or more individual things, are relational properties, or relations.

 

 

We can define property constants so that they represent relational properties, e.g.,

 

Txy = x is taller than y

Sxy = x is shorter than y

Mxy = x is the mother of y

Cxy = x is the child of y

 

So if we let “a” stand for Andrew and “b” stand for Bill…

 

Tab      =          Andrew is taller than Bill.

 

And notice that the order of “a” and “b” makes a big difference. If “Txy” has been defined as “x is taller than y,” then

 

Tba      =          Bill is taller than Andrew.

 

Another example:

 

Sxyz    =          x sat between y and z

 

So letting “c” stand for Cindy…

 

Sacb     =          Andrew sat between Cindy and Bill

Sbac     =          Bill sat between Andrew and Cindy

Scab     =          Cindy sat between Andrew and Bill

 

 

[8.1.2.] Relational Property Constants and Quantifiers.

 

Relational predicates can be integrated with quantifiers:

 

“Everyone knows Bill.”

 

To symbolize this sentence, use the following strategy:

 

 

 

1.      Ask to whom or what the sentence is attributing a property:

 

This sentence is attributing a property to all people, and this suggests that the symbolization will look like this:

 

(x)(Px É ...)    [Px: x is a person]

 

2.      Ask what property is being attributed:

 

In this case, the property that is being attributed is the property of knowing Bill.

 

The sentence means: for all x, if x is a person, then x knows Bill.

 

So we will need a relational predicate for knowing:           Kxy: x knows y

 

And we will need an individual constant to refer to Bill: b

 

The sentence should thus be symbolized:

 

(x)(Px É Kxb)       [Kxy: x knows y]

 

 

Other examples [all with unrestricted domains of discourse]:

 

“Bill knows everyone.”                                               (x)(Px É Kbx)

“Bill knows someone.”                                                ($x)(Px · Kbx)

“Someone knows Bill.”                                               ($x)(Px · Kxb)

“If everyone knows Bill, then someone does.”            (x)(Px É Kxb) É ($x)(Px · Kxb)

“Bill doesn’t know anybody.”                         ~($x)(Px · Kbx) or (x)(Px É ~Kbx)

 

[the following have a domain of discourse restricted to all people]

 

“Bill knows everybody Andy knows”             (x)(Kax É Kbx)              or   ~($x)(Kax · ~Kbx)

“Andy knows everybody Bill knows”             (x)(Kbx É Kax)   or   ~($x)(Kbx · ~Kax)

“Andy knows somebody Bill doesn’t know”   ($x)(Kax · ~Kbx)

 

 

Exercise 10-1 (p.228)

·         do all problems for next time; we’ll review the odds next class

·         notice that the book uses the same letter to represent a one-place property and a two-place property in a single sentence (e.g. #3... Mxy = x is married to y, and Mx = x is married).

·         notice that in #20 there is a 4-place predicate!

 

 

 

[10.2.] “Symbolizations Containing Overlapping Quantifiers.” (Ch. 10:2)

 

Recall that the scope of a quantifier is the part of a sentence in which variables may be bound to that quantifier. For example...

 

(x)Fx                                        scope of “(x)” = “Fx”

(x)(Fx É Gx)                            scope of “(x)” = “(Fx É Gx)”

(x)(Fx É Gx) Ú Gx                   scope of “(x)” = “(Fx É Gx)”  [scope does not include “Gx” – so either the third “x” is a quasivariable (which is possible if this sentence occurs in a proof), or this is a sentence form instead of an actual sentence]

 

It is possible for an expression to contain two or more quantifiers the scopes of which overlap.

 

E.g. (these examples come from your textbook-- p.229)

 

[to keep these examples simple, I am limiting the domain of discourse to persons]

 

Everyone loves everyone.       (x)(y)Lxy

 

I.e., for any two members of the DOD you happen to choose, one loves the other.

 

Someone loves someone.        ($x)($y)Lxy

           

I.e. there is at least one individual in the DOD that loves at least one individual in the DOD.

 

Not everyone loves everyone. ~(x)(y)Lxy   or   ($x)($y)~Lxy

 

I.e., it is not the case that for any two members of the DOD you happen to choose, one loves the other; i.e., there are at least two members of the DOD about whom it is the case that one does not love the other.

 

Notice that the formulae   ~(x)(y)Lxy   and   ($x)($y)~Lxy   are equivalent; you can get from one to the other by two applications of the equivalence rule QN:

 

1. ~(x)(y)Lxy               p                                  1. ($x)($y)~Lxy           p

2. ($x)~(y)Lxy             1 QN                            2. ($x)~(y)Lxy             1 QN

3. ($x)($y)~Lxy           2 QN                            3. ~(x)(y)Lxy               2 QN

 

No one loves anyone.              ~($x)($y)Lxy  or   (x)(y)~Lxy

 

I.e., it is not the case that there is at least one individual in the DOD that loves at least one individual in the DOD; i.e., for any two members of the DOD you happen to choose, it is not the case that one loves the other.

 

Notice that the formulae   ~($x)($y)Lxy  and   (x)(y)~Lxy   are equivalent; you can get from one to the other by two applications of the equivalence rule QN:

 

1. ~($x)($y)Lxy           p                      1. (x)(y)~Lxy               p

2. (x)~($x)Lxy             1 QN                2. (x)~($y)Lxy             1 QN

3. (x)(y)~Lxy               2 QN                3. ~($x)($y)Lxy           2 QN

 

 

Exercise 10-2 (pp.231-32)

·         ignore the second half of the instructions (“...and construct their expansions in a two-individual universe of discourse.”) We’re skipping all the sections on expansions, like sec. 10:3.

·         We did 2, 4, 6, and 8 in class; do all of the rest for next time; we’ll check the odds in class next time.

 

Exercise 10-3, pp.233-34

·         do the all of these; we’ll check the odds next time

·         look out for #17 -- it’s relatively difficult

 

 

[8.3.] “Places and Times.” (Ch.10:4)

 

We can use the notation of relational predicate logic to translate natural language sentences that refer to places. An example from your textbook (234):

 

“Somewhere, the streets are all paved with gold.”

 

A first step is to recognize that the use of “somewhere” indicates that the sentence is claiming that a specific sort of thing exists, namely, a place:

 

There exists a place where all streets are paved with gold.

 

Px: x is a place

Sxy: x is a street in y

Gx: x is paved with gold

 

($x)[Px · (y)(Syx É Gy)]

 

 

We can also translate sentences that refer to times into the symbolization of relational predicate logic:

 

“The sun never sets on the British Empire.”[1]

 

Similarly to the previous example, a good first step is to recognize that the use of the word “never” indicates that the existence of a certain sort of thing is being denied: the sentence is denying the existence of a certain sort of time:

 

There is no such thing as a time at which the sun sets on the British Empire.  =

 

Tx: x is a time

Sxyz: x sets on y at z

s: the sun

b: the British Empire

 

~($x)(Tx · Ssbx)

 

 

Exercise 10-4, pp.234-35

·         we did #2 and 4 in class; do all of the rest of these for next time; we’ll work through the odds in class.

 

 

Stopping point for Tuesday April 10. For next time:

·         complete ex. 10-1, 10-2, 10-3, and 10-4;

·         read Ch.10:5, 7-8 (pp.235-38, 241-49).