Relational Predicate Logic - The University of West Georgia

Symbolic Logic (PHIL 4160)
Robert Lane, UWG
April 8, 2014
[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”
“Dumbo is an elephant”
“Someone is conceited”
“All cows are mammals”
=
=
=
=
Tm
Ed
(∃x)(Px • Cx)
(x)(Cx ⊃ 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. It makes 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
Sbac
Scab
=
=
=
Andrew sat between Cindy and Bill
Bill sat between Andrew and Cindy
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 what or to whom the sentence is attributing a property:
This sentence is attributing a property to all people, and this suggests that the symbolization – in an
unrestricted domain of discourse – will look like this:
(x)(Px ⊃ ...) [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 ⊃ Kxb)
[Kxy: x knows y]
Other examples [all with unrestricted domains of discourse]:
“Bill knows everyone.”
“Bill knows someone.”
“Someone knows Bill.”
“If everyone knows Bill, then someone does.”
“Bill doesn’t know anybody.”
(x)(Px ⊃ Kbx)
(∃x)(Px • Kbx)
(∃x)(Px • Kxb)
(x)(Px ⊃ Kxb) ⊃ (∃x)(Px • Kxb)
~(∃x)(Px • Kbx) or (x)(Px ⊃ ~Kbx)
[the following have a domain of discourse restricted to all people]
“Bill knows everybody Andy knows”
“Andy knows everybody Bill knows”
“Andy knows somebody Bill doesn’t know”
(x)(Kax ⊃ Kbx) or ~(∃x)(Kax • ~Kbx)
(x)(Kbx ⊃ Kax) or ~(∃x)(Kbx • ~Kax)
(∃x)(Kax • ~Kbx)
Exercise 10-1 (p.232-233)
• write down all odd 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
(x)(Fx ⊃ Gx)
(x)(Fx ⊃ Gx) ∨ Gx
scope of “(x)” = “Fx”
scope of “(x)” = “(Fx ⊃ Gx)”
scope of “(x)” = “(Fx ⊃ 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]
In order to symbolize some sentences containing predicates of two or more places, we have to use
two or more quantifiers the scopes of which overlap.
E.g. (these examples come from your textbook-- p.233)
[to keep these examples simple, I am limiting the domain of discourse to human beings]
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.
(Notice that this can be true even if there is only one human being in the domain of discourse…
so long as that human loves himself or herself.)
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
2. (∃x)~(y)Lxy
3. (∃x)(∃y)~Lxy
No one loves anyone.
p
1 QN
2 QN
1. (∃x)(∃y)~Lxy
2. (∃x)~(y)Lxy
3. ~(x)(y)Lxy
p
1 QN
2 QN
~(∃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
2. (x)~(∃x)Lxy
3. (x)(y)~Lxy
p
1 QN
2 QN
1. (x)(y)~Lxy
2. (x)~(∃y)Lxy
3. ~(∃x)(∃y)Lxy
p
1 QN
2 QN
Exercise 10-2 (pp.236)
• for next time, write answers to all odd-numbered problems
• 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.
Exercise 10-3, pp.237-38
• write down answers to all odd-numbered problems
• 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 (p.238):
“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 ⊃ 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
• write down answers to the odd-numbered problems
Stopping point for Tuesday April 8. For next time:
• complete the odd-numbered problems in: ex. 10-1, 10-2, 10-3, and 10-4 (#5-15 only, assuming
that we finished 1 through 4 just now);
• read Ch.10:5, 7-8 (pp.239-242, 246-254)
1
This example is from a previous edition of your textbook, see p.238 for the new version’s (overtly political)
example, which involves the “U.S. Empire.” The author’s comment that this is a “dated example” is actually left
over from that previous edition and so presumably refers to the British Empire example.

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