17.
INFERENCE RULES FOR
QUANTIFIERS
The final problem in your
last set of exercises called for working with the set of inference and
equivalence rules developed so far, but clearly we need more rules in
order to handle the type of syllogism I proposed early on as an
example of a valid argument that could still have a false conclusion
since one or more premises are also false.
Everything black is sweet, and
salt is black, so salt is sweet.
Symbolization is not the
problem: (x)(Bx -> Wx),
(x)(Sx -> Bx) |- (x)(Sx -> Wx)
Note that since
both "sweet" and "salt" begin with the same letter, we will let "W"
represent the predicate of being sweet.
You should see the parallel with our
hypothetical syllogisms from propositional logic (P->Q, Q->R |- P->R),
but we cannot directly apply the rule here since the quantifiers and
not the arrows are the main connectives in the three expressions.
What we can do is instantiate each expression--meaning, we apply the
generalization to just one individual--and after that we do work with
our usual inference rules, then we again generalize so that once more
we say it is true of every individual (each spoonful of salt, let's
say) in our category. The new rules we will work with are called universal instantiation (UI) and universal generalization (UG).
(x)Fx |- Fa
UI (and we may use any letter apart from our
variables of x and y and z)
Note that, unlike many texts, including
Zegarelli's, I do not allow the use of "free" variables (for
instance, Fx by itself)
|
(x)(Fx -> Gx) Fa -> Ga
UI
Fa |- (x)Fx UG (but here we limit the rule to a
situation in the name used is "free" in the sense that originally it
could have been true of anyone at all)
Fa -> Ga |- (x)(Fx ->
Gx) UG
An "illegal" use of the rule would be in
a derivation for an argument form in which there had not been a way in
which we could say everyone would have the characteristic in
question. For instance, if the premises include both (x)(Fx -> Gx) and Fa we can certainly say Fa -> Ga, but even knowing
that Ga would follow we
are not able to say (x)Gx.
The reason is that we are saying any individual who has the
characteristic F will be
one of those who do have the characteristic G, but this will not justifiy
saying everyone will have that characteristic.
Later we will see an important exception in which we will go from Fa -> Ga to (x)(Fx -> Gx).
We apply the same instantiation
technique to existential expressions but with a very crucial
catch. With a universal quantifier we can see how it would not
matter what name we use when we cite a particular instance. With
an existential quantifier we are saying that there has to be at least
one individual with that characteristic, but we cannot make the mistake
of saying it has to be someone we would already be talking about.
This is often called the new
name rule, and we will build it into the application of existential instantiation by
making sure that the name has not already be in use. What this
means in practice is that in a proof we instantiate an existential
quantifier before we instantiate a universal quantifier, since our
universal quantifier can already apply to our hypothetical individual a.
(Ex)Fx |- Fa EI
(the name used here must always be a "new" name, one not appearing in
the argument form or in a previous step in the proof.
(Ex)(Fx & Gx) |- Fa &
Ga EI
For existential generalizations
we have no problem at all. Saying something is true of any
individual means that it must be true of some individual.
Fa |- (Ex)Fx EG
Fa & Ga |- (Ex)(Fx & Gx) EG
Note that with our two generalization rules we might have other
connectives inside the parenthesis, but most typically we are working
with the arrow when we have universal quantifiers and ampsersands
when we have existential quantifiers. The reason is that in
symbolizing an expression such "all students are ambitious" we are
rethinking it in terms of a conditional ("if x is a student then x is
ambitious"), while the expression "some students are ambitious' (or
"there is an ambitious student") asks us to state two things about x: x
is a student and x is ambitious.
DERIVATIONS
Whenever the quantifier itself is the main connective, we cannot work
with any of our previous inference rules (although we may always work
with our replacement rules) until we have eliminated the
quantifier. Sometimes the letter used ("a" or "b" or whatever)
represents an actual name, sometimes a hypothetical one (always the
case with the EI rule). Then we proceed just as we have
before. Note that for the present we are working only with what
are called direct derivations, which will not apply in all cases.
Shortly we will be learning another technique that does.
Everybody
who is
ambitious will work
hard.
Some students are ambitious. Therefore, some students will work
hard.
(x)(Ax -> Wx), (Ex)(Sx & Ax) |- Ex(Sx
&
Wx)
1.
(x)(Ax
-> Wx)
2. (Ex)(Sx & Ax)
/ show (Ex)
(Sx & Wx)
3. Sa &
Aa
2, EI note that
we have to do this step first in order to have a new name
4. Aa ->
Wa
1, UI
5.
Aa
3, Simp
6
Wa
4,5, MP
7.
Sa
3, Simp
8. Sa &
Wa
6,7, Conj
9. (Ex)(Sx & Wx)
8, EG
We could, if we wish, use different
variables in our symbolization so that we could begin with one and end
up with another, as in the example below.
Everyone's
bright and everyone's charming. Therefore, everyone is both
bright and charming.
(x)Bx
&
(y)Cy
|- (z)(Bz & Cz)
1.
(x)Bx
& (y)Cy
/ show (z)(Bz & Cz)
2.
(x)Bx
1, Simp note that since the main connective
is
"&"
we must eliminate this before we go ahead
3.
Ba
2, UI
4.
(y)Cy
1, Simp
5.
Ca
4, UI
6. Ba
&
Ca
3,5, Conj
7.
(z)(Bz &
Cz)
6, UG
EXERCISES (ON YOUR OWN)
Symbolize and provide derivations
for each of the following.
1. Every student is
ambitious. No one who is ambitious will be lazy. Therefore,
there will be no lazy students.
2. Some students do work
hard. Anyone who works hard will pass. Therefore, there are
students who will pass.
3. Alice is an ambitious
student and Bob is a bright student. If there are both bright
students and amitious students, then the teacher will be happy.
No teachers, then, will be unhappy.
4. Some classes are not
interesting. Any class that is not interesting will not be
fun. If there are classes that are not fun then Bob will not be
happy. It does follow that Bob will be unhappy.
5.
Alice will pass only if all the classes are interesting.
Only classes
that are fun are interesting. Some classes are not fun, so it
follows
that Alice is not going to pass.
Go to the answer key
What we are calling
the new name rule is particularly important. Failing to respect
it is one of the two most serious mistakes made by even very good
students (the other is failure to work just with the main connective in
using any inference rule).