[8.] Predicate Logic Proofs.

 

 

[8.1.] “Proving Validity.” [Ch. 9:1]

 

Just as in sentential logic, in predicate logic it is possible to use the proof method to demonstrate that an argument is valid. For example:

 

            1. (x)(~Fx Ú Gx) É ($x)Hx                    p

2. (x)(~~Fx É Gx)                                 p          /\ ($x)Hx

3. (x)(Fx É Gx)                                     2 DN

4. (x)(~Fx Ú Gx)                                   3 Impl

5. ($x)Hx                                              1, 4 MP

 

This example illustrates that in predicate logic (just as in sentential logic), equivalence rules (like DN and Impl) can be applied to parts of lines.

 

And just as in sentential logic, implicational rules (like MP) can be applied only to entire lines…

 

So, although the following argument is valid...

 

1. (x)(Fx É Gx)

2. (x)Fx                        /\ (x)Gx

 

this “proof” does not show it to be valid:

 

1.       (x)(Fx É Gx)           p

2.       (x)Fx                      p          /\ (x)Gx

3.       (x)Gx                      1, 2 MP

 

This “proof” fails because it applies MP to part of premise one, not to the entire line. Premise one does not have the form p É  q , so MP cannot be applied to it. (It’s actual form is simply p. In so far as MP is concerned, the first premise may as well be an atomic sentence.)

 

In order to prove this sort of argument valid, we need new rules: four implicational rules and one equivalence rule.

 

About the four implicational rules...

·         two of them are for removing quantifiers from lines

·         two of them are for introducing quantifiers into lines

 

The general strategy we will use:

·         step one: remove quantifiers from premises

·         step two: use original rules (eight implicational rules, ten equivalence rules, CP and IP) to derive new lines

·         step three: add quantifiers to those new lines (to derive the conclusion of the argument)

 

 

[8.2.] “The Four Quantifier Rules.” [9:2]

 

See the inside cover of your textbook for summary statements of these new rules. They are not as easy to state as the original rules, and each of them takes some detailing explanation.

 

 

[8.2.1.] Universal Instantiation (UI).

 

This rule allows you to move from a universal quantification to an instance of that quantification (hence the name “universal instantiation”).

 

The rule may be stated:

·         “From any universal quantification we may validly infer any instance of it.”[1]

·         “[Universal i]nstantiation is an operation that consists in deleting a quantifier and replacing every variable bound by that quantifier with the same instantial letter.”[2]

 

The general idea behind this rule is: whatever is true of everything is true of any particular thing.

 

For example:                 Ex = x is an elephant

Mx = x is a mammal

d = Dumbo

 

1. (x)(Ex É Mx)                                    p

2. Ed                                        p          /\ Md

 

(UI) allows you to derive the first of these two lines from line 1:

 

3. Ed É Md                               1 UI

4. Md                                       2, 3 MP

 

After all, line 1 asserts that for any member of the domain of discourse (and as before, we are assuming an unrestricted domain of discourse), if that thing is an elephant, then it is a mammal. It follows from this that if Dumbo is an elephant, Dumbo is a mammal.

 

From the point of view of simply applying UI correctly, it doesn't matter which individual you refer to in the instantiation, and thus it does not matter which individual constant you use. You could have also inferred Ea É Ma, Eb É Mb or any other instantiation of (x)(Ex É Mx). If it is the case that, for anything, if it is an elephant then it is a mammal, then it is the case that, for any particular thing you choose, if it is an elephant then it is a mammal.

 

 

[8.2.1.1.] Deciding Which Instance to Infer.

 

An important question of strategy: How do you decide which instance of the universal quantifier to infer?

 

In deciding on an instance, you should keep in mind the conclusion you are trying to reach, as well as what constants are used in other premises. In the above argument about Dumbo (“d”), it would have done you no good to use a constant other than “d”. Deriving “Ea É Ma” from line one would have been a valid move, but it would have been bad strategically, since it would not have helped you solve the proof.

 

Here is another example that illustrates this:

 

1. (x)(~Hx É Ix)                       p

2. (y)(~Ky Ú ~Iy)                      p

3. ~Hg                                      p          / \ ~Kg

 

The rule UI will allow you to use any constant you like in instantiating the universal quantifications on lines 1 and 2, but strategically it makes sense to use only the constant “g”:

 

4. ~Hg É Ig                              1 UI

5. ~Kg Ú ~Ig                             2 UI

6. Ig                                         3, 4 MP

7. ~~Ig                                     6 DN

8. ~Kg                                      5, 7 DS

 

 

[8.2.1.2.] Two More Things to Note About UI.

 

One

 

UI is an implicational rule. So the quantification that is replaced must be an entire line in the proof. This means that you can apply this rule to a line only if the entire line is a universal quantification.

·         You cannot apply UI to (x)Fx É (x)Gx, which is a conditional, not a universal quantification.

·         Nor can you apply it to (x)Fx Ú (x)Gx, which is disjunction.

·         You cannot even apply it to ~(x)(Fx É Gx), which is a negation.

 

Two

 

When using UI, you can replace the variable with either an individual constant (which stands for some definite individual) or a quasivariable:

 

quasivariable (df.) (a.k.a. an unknown): a lower case letter (x, y, z…) that does not work as a variable but instead refers to some indefinite individual, i.e. to some individual the identity of which is unknown. It is like a marker that refers to an individual that has not been definitely identified.

 

So in the above proof, a valid move at line 3 would have been: “Ey É My” (such a move would have been bad strategically, for the same reason that using a constant other than “g” would have been bad: it would not have helped you solve the proof. But it would nonetheless be a valid move). Even though “y” is not bound to a quantifier, this is still a sentence rather than a sentence form. It is a quasivariable, an ambiguous name that works more like a free variable than like a constant. We will see why we want to do this when we examine the next two rules...

 

 

[8.2.2.] Universal Generalization (UG).

 

This rule allows you to move from an instance of a universal quantification back to the quantification itself:

 

Ex = x is an elephant

Mx = x is a mammal

Ax = x is an animal

 

1. (x)(Ex É Mx)                                    p

2. (x)(Mx É Ax)                       p          /\ (x)(Ex É Ax)

 

Our strategy will be:

·         use UI to remove the quantifiers from lines 1 and 2 (since there are no constants already in the proof, we will instantiate using a quasivariable)

·         apply the rule HS

·         use UG to introduce the quantifier back into the proof, thus arriving at the desired conclusion.

 

3. Ex É Mx                         1 UI

4. Mx É Ax                                    2 UI

5. Ex É Ax                         3, 4 HS

6. (x)(Ex É Ax)                  5 UG

 

There are important constraints on the use of this rule that we will cover next time.

 

 

[8.2.3.] Existential Instantiation (EI).

 

This implicational rule allows you to move from an existential quantification to an instance of that quantification (hence the name “existential instantiation”). For example:

 

1. ($x)Dx          Something is a duck.

 

We can use the rule EI to eliminate the quantifier from that premise:

 

2. Dy                y is a duck.

 

Here, “y” is a quasivariable, an ambiguous name. It refers to a specific individual, but it doesn’t say which individual (Daffy, Donald, the AFLEC duck, etc.)

 

There are two restrictions on the use of EI – you must learn these in order to use this rule correctly:

 

EI RESTRICTION #1

 

The variable in the existential quantification must be replaced with a quasivariable.

 

So the following would be invalid:

 

1. ($x)Px                      p

2. Pd                            1 EI

 

 

The fact that that there is a philosopher doesn’t not imply that any particular thing is a philosopher (here “d” stands for Dumbo the elephant; so this is clearly invalid).

 

A proper application of EI is as follows:

 

1. ($x)Px                      p

2. Px                            1 EI

 

 

 

EI RESTRICTION #2

 

The quasivariable must not appear free (unbound) in any line already in the proof.

 

The following “proof” would violate this second restriction... (Fx = x is a frog)

 

 

1. ($x)Dx · ($x)Fx              p

2. ($x)Dx                            1 Simp

3. ($x)Fx                            1 Simp

4. Dy                                  2 EI

 

Up to here, each move has been valid. But if we continue as follows...

 

5. Fy                                   3 EI

 

we’ve done something illegitimate. This is because “y” on line 5 stands for the same (unknown) individual as “y” stands for on line 4. Line 4 asserts that y is a duck; line 5 asserts that y is a frog. Clearly, from the fact that something is a duck and something is a frog (which is asserted in line 1), it does not follow that there is one and the same thing, y, that is both a duck and a frog.

 

So when you apply EI, you must choose a letter for your quasivariable that has not appeared free (unbound) in the proof up to that point. So we could use either “x” or “z”...

 

5. Fz                                   3 EI (this is OK because “z” hasn’t yet appeared in the proof)

 

or

 

5. Fx                                   3 EI      (this is OK because “x” is bound in 1 and 2)

 

 

 

 

[8.2.4.] Existential Generalization (EG).

 

This rule may be the easiest of all of the new rules to understand. It is based on the simple idea that if some specific individual has a property, then it is true that there is something that has that property. For example,

 

1. Ed                p

2. ($x)Ex          1 EG

 

As with the other rules for adding and removing quantifiers, EG can only be applied to entire lines. So the following argument misapplies this rule:

 

            Mx = x is a mouse

Dx = x is a dog

m = Mickey

p = Pluto

 

1.       Mm · Dp                            p          /\ ($x)Mx · ($x)Dx

2.       ($x)Mx · ($x)Dx                 1 EG     invalid!

 

To capture the validity of this argument in a proof, you would need to do the following:

 

1. Mm · Dp                              p          /\ ($x)Mx · ($x)Dx

2. Mm                                      1 Simp

3. Dp                                        1 Simp

4. ($x)Mx                                 2 EG

5. ($x)Dx                                  3 EG

6. ($x)Mx · ($x)Dx                   4, 5 Conj

 

 

Exercise 9-1 (p.205)

·         Complete all problems for next time; we’ll check the odds at the start of class

·         Remember that you can use all 18 implicational and equivalence rules, as well as CP or IP.

 

 

Stopping point for Monday March 24. For next time: exercise 9-1, and read Ch.9:3 (pp.205-209).