Proving Invalidity

1
Proving Invalidity

• Recall that natural deduction (the 18 rules we
  learned in chapter 7) allows us to prove certain
  arguments valid without using truth tables. But
  it is also true that our failure to prove an
  argument valid does not automatically mean the
  argument is invalidperhaps we just were not
  clever enough in using the rules
• However, we could always use truth tables to show
  for certain whether an argument was valid or
  invalid

2
Not So Easy with Predicate Logic

• All humans are mortal
• Socrates is human
• Therefore, Socrates is mortal
• This is valid, but we cannot prove it so using
  truth tables. To see this, note that if we
  assigned each statement a sentence letter (e.g.
  H, S, and M, respectively), then there would be a
  row of the truth table where H and S are true,
  but M false

3
And we Cannot Represent it as Modus Ponens

• All humans are mortal
• Socrates is human
• Therefore, Socrates is mortal
• We cannot represent 1 as a conditional, 2 as the
  antecedent of that conditional, and 3 as the
  consequent. For then the conditional in 1 would
  have to be represented as If Socrates is human,
  then Socrates is mortal. But 1 is saying
  something about all humans, not just Socrates

4
What to Do?

• Since we cannot always use truth tables to prove
  the validity or invalidity of arguments in
  predicate logic, we need other methods to show an
  argument is invalid.
• 2 methods
• Counterexample method
• Finite universe method

5
Counterexample Method

• We talked about this method early in the term
• The trick is to come up with an interpretation of
  a symbolized argument that makes the premises
  clearly true and the conclusion clearly false.
• Example
• (?x)(Dx Cx)
• (?x)(Dx Gx) / (?x)(Cx Gx)

6
See?

• (?x)(Px Cx)
• (?x)(Px Dx) / (?x)(Cx Dx)
• Let P is a pet, C is a cat, and
• D is a dog
• The first premise says that there is something
  that is a pet and a cat (true), the second
  premise says there is something that is a pet and
  a dog (true), but the conclusion says there is
  something that is a cat and a dog (false). So the
  argument is invalid

7
Relational Predicates

• The predicates we have used to this point have
  assigned attributes to single things
• Examples Lassie is a dog (Dl), Joe is a
  student (Sj)
• Such predicates are called monadic, or
  one-place
• But predicates can also be used to express
  relations between or among things
• Examples Henry is the brother of Steve (Bhs),
  Chicago is between San Francisco and New York
  (Bcsn)
• The first is an example of a two-place
  predicate, the second an example of a
  three-place predicate
• Note that in multi-placed predicates, the order
  of the lower case letters following the predicate
  letters can affect the meaning of the statement

8
Overlapping Quantifiers

• To this point, the quantifier expressions we have
  considered have not had quantifier expressions
  among their components
• Examples
• (x) (Hx?Jx)
• This is a universally quantified claim that has
  no quantifier expressions as components
• 2. (?x)(Dx Cx)
• This is an existentially quantified claim that
  has no quantifier expressions as components

9
But Consider

• Every cat hates every dog
• This statement contains two quantifiers, and they
  overlap that is, it is saying something about
  all cats and all dogs and how they are related
• How should we represent what this says?
• (x)(Cx?(y)(Dy?Hxy)
• This is a universally quantified claim that has a
  universally quantified claim as a component.
• Notice that the x in Hxy is bound by an
  x-quantifier and that the y in Hxy is bound
  by a y-quantifier

10
Note

• Imagine we are talking just about people
• 1. Everyone loves someone
• (x)(?y) Lxy
• 2. Someone loves everyone
• (?x)(y) Lxy
• These say different things!
• 1. says that each person loves at least one
  person 2. says there is at least one person who
  loves every person
• So when we are considering statements with
  overlapping quantifiers that differ in kind
  (existential and universal), the order of the
  quantifiers matters to what is being asserted
  otherwise, the order does not matter

11
Using Variables

• When symbolizing claims with overlapping
  quantifiers, choose different variables to avoid
  ambiguity
• Example
• 1. (x)Gx (y)(Gy?Sy)?Sx
• Vs.
• 2. (x)Gx (x)(Gx?Sx)?Sx
• In 2, the variables in Gx?Sx appear to be bound
  by 2 quantifiers. They can be bound by only one,
  however, so 1 is the appropriate symbolization

12
Lets Translate!

• Every person can sell something or other
• Steve invited some of his friends
• Every person admires some people he or she meets
• Some policemen arrest every traffic violator they
  see

13
See?

• Every person can sell something or other
• (x)Px?(?y) Sxy)
• 2. Steve invited some of his friends
• (?y)(Fys Isy)
• 3. Every person admires some people he or she
  meets
• (x)Px?(?y)Py (Mxy?Axy)
• Some policemen arrest every traffic violator they
  see
• (?y)Py (x)(Tx Syx)?Ayx

14
Applying the Rules of Inference

• The change of quantifier rule applies in the same
  way
• The rules for adding and dropping quantifiers
  apply in the same way, with one exception
• UG cannot be used if Fy contains a constant and y
  is free in the line where that constant is
  introduced

15
Why Not?

• Remember that everyone loves someone and
  someone loves everyone say different things,
  and the former does not imply the latter.
  Without the further restriction on UG, however,
  we could derive the latter from the former.

16
Like This

• (x)(?y) Lxy
• (?y)Lzy 1, UI
• Lza 2, EI
• (x)Lxa 3, UG
• (?y)(x)Lxy 4, EG
• The mistake is on line 4 we cannot apply UG to
  line 3, since a is introduced on that line and
  z occurs free on that line

17
More on Choosing Variables

• When you apply UI and decide to use a variable as
  the instantial term, you must make sure the
  variable you choose winds up bring free
• (x)(y)Lxy
• (y)Lyy 1, UI Mistake!
• In dropping the x-quantifier and replacing x with
  y, the instantial term does NOT end up free, but
  is bound by the y-quantifier

18
Continued

• When applying UG and EG, make sure that the
  variable you choose is not already bound by a
  previous quantifier, and make sure no additional
  variables end up bound as a result of
  generalizing
• Example
• Lyd
• (?y)Lyy 1, EG Mistake!
• In this case, by choosing y to replace d, we
  ended up binding the free y in 1.

19
Another Example

• (?y) Lxy
• (y)(?y) Lyy 1, UG Mistake!
• In applying UG to 1, we choose a variable (y)
  that was already bound by another quantifier

20
Lets Try It Out

• Any professional can outplay any amateur. Jones
  is a professional but he cannot outplay Meyers.
  Therefore, Meyers is not an amateur.
• Symbolize this argument and prove that it is valid

21
Symbolizing

• Any professional can outplay any amateur. Jones
  is a professional but he cannot outplay Meyers.
  Therefore, Meyers is not an amateur.
• (x)Px?(y)(Ay?Oxy)
• Pj Ojm /Am
• Pj?(y)(Ay?Ojy) 1, UI
• Pj 2, Simp
• (y)(Ay?Ojy) 3, 4 MP
• Am?Ojm 5, UI
• Ojm Pj 2, Com
• Ojm 7, Simp
• Am 6, 8 MT

22
One More

• OBrien is a person. Furthermore, OBrien is
  smarter than any person in the class. Since no
  person is smarter than himself, it follows that
  OBrien is not in the class
• Symbolize and prove the argument valid

23
Symbolize

• OBrien is a person. Furthermore, OBrien is
  smarter than any person in the class. Since no
  person is smarter than himself, it follows that
  OBrien is not in the class
• Po
• (x)(Px Cx)?Sox
• (?y) (Py Syy) /Co

24
Prove

• Po
• (x)(Px Cx)?Sox
• (?y) (Py Syy) /Co
• Co AIP
• Po Co 1, 4 Conj
• (Po Co)?Soo 2, UI
• Soo 5, 6 MP
• Po Soo 1, 7 Conj
• (y)(Py Syy) 3, CQ
• (Po Soo) 9, UI
• (Po Soo) (Po Soo) 8, 10 Conj
• Co 4-11 IP