### PHIL 4160: Symbolic Logic Dr. Robert Lane Lecture Notes: Thursday April 12, 2012

[8.4.] “Symbolizing ‘Someone’, ‘Somewhere’, ‘Sometime’, and So On.” (Ch.10:5)

Some uses of the terms “someone,” “somewhere,” etc., are misleading. At this point, you are probably tempted to reach for an existential quantifier — “(\$x)” — anytime you see these words. But some sentences that use these words are better symbolized using a universal quantifier: “(x)”.

For example:

“If someone is late, then everyone is annoyed with that person.”

You may be tempted to symbolize this sentence as follows:

dod = persons

(\$x)[Lx É (y)Ayx]

This says: there is at least one person for which the following is true: if he or she is late, then everyone is annoyed with him or her. But this sentence is true if there is only one person in the world who would annoy everyone by being late. E.g., this sentence would be true if Clark would annoy everyone by being late but Lois would not (maybe everyone likes Lois more and so is less annoyed by her tardiness).

This is not what our original sentence means. Our original sentence, even though it uses the word “someone,” is attributing a property to every individual in the domain of discourse (viz., that he or she will make everyone annoyed if he or she is late). So we should use a universal quantifier:

(x)[Lx É (y)Ayx]

Important points to notice:

·         The mistake of using “(\$x)” rather than “(x)” is another example of incorrect combination of existential quantifier and horseshoe. Remember: if your symbolization is an existential quantification of a conditional, you should be suspicious. It is very unusual for a sentence to be accurately symbolized in that way, just as it is unusual for a sentence to be symbolized as a universal quantification of a conjunction, as in “(x)(Fx · Gx)”.

·         The symbolizations

(\$x)Lx É (y)Ayx

and

(x)Lx É (y)Ayx

would also be incorrect, since in each one, the second “x” is a free variable, i.e., a variable not bound by a quantifier. This is because the second “x” falls outside the scope of the “x” quantifier.

·         The sentence “If someone is late, then everyone is annoyed,” if interpreted strictly (to mean that if any person anywhere is late, then everybody everywhere will be annoyed) can be symbolized with an existential quantifier:

(\$x)Lx É (y)Ay                 which is equivalent to              (x)[Lx É (y)Ay]

A more complicated example (from p.232):

"If someone is too noisy, then if everyone in the room is annoyed, someone will complain."

We can partially translate this correctly with an existential quantifier:

If (\$x)Nx then, if (y)(Ry É Ay) then (\$z)Cz

and then complete the translation as:

(\$x)Nx É [(y)(Ry É Ay) É (\$z)Cz]

But the following cannot be translated that way:

"If someone is too noisy, then if everyone in the room is annoyed, someone will complain about that person."

We cannot translate this sentence as:

(\$x)Nx É [(y)(Ry É Ay) É (\$z)Czx]

because the final x is unbound, i.e., it is a free variable, not within the scope of the “x’ quantifier.

And we cannot solve the problem by bringing the “x” within the scope of the “(\$x)” with braces:

(\$x){Nx É [(y)(Ry É Ay) É (\$z)Czx]}

…because what that symbolization says is: there is at least one person about whom the following is true: if he or she is late and everyone in the room is annoyed, someone will complain about him or her. And this is consistent with there being only one person whose tardiness would cause annoyance and a complaint, as in the earlier Clark and Lois example (again, notice the suspicious pairing of the existential quantifier with a conditional).

Our original sentence does not mean that there is at least one such person. Rather, it means that anyone whose tardiness causes everyone in the room to be annoyed will be complained about by someone. So as with the first example, we need to use a universal quantifier:

(x){Nx É [(y)(Ry É Ay) É (\$z)Czx]}

There are more examples on the box on pp.237-38.

Exercise 10-6 (p.239)

·         complete all for next time; we will cover at least the odds in class

Exercise 10-7 (p.239-40)

·         [we will do at least the first four in class]

·         complete all for next time; we will cover at least the odds in class

[8.5.] Relational Predicate Logic Proofs. (Ch.10:7)

We do not need to add any additional rules in order to construct proofs in relational predicate logic. The proofs in this section can be constructed using only the following:

·         the original eight implicational rules (MP, MT, DS, Simp, Conj, HS, Add, CD)

·         the original ten equivalence rules (DN, DeM, Comm, Assoc, Dist, Contra, Impl, Exp, Taut, Equiv)

·         conditional proof (CP)

·         indirect proof (IP)

·         the four implicational rules for quantifiers (UI, UG, EI, EG)

·         quantifier negation (QN, the only equivalence rule for quantifiers)

However, some of the rules introduced for predicate logic proofs have more restrictions when used in relational predicate logic.

[8.5.1.] Universal Instantiation.

Something new in these proofs is that you will have premises containing quantifiers with overlapping scopes. For example,

Fxy = x is the father of y

Cxy = x is the child of y

1. (x)(y)(Fxy É Cyx)               p

2. Fab                                      p          /\ Cba

As before, your strategy should be to use UI in order to remove the universal quantifiers so that you can apply other rules to finish the proof.

What is new is that you will need to apply UI more than once in order to remove multiple quantifiers from a given premise. You will apply it one time to remove the first quantifier (thus generating a new line in the proof), then apply it a second time (to the newly generated line) to remove the second quantifier.

Since UI can be applied only to an entire sentence (in particular, it can be applied to a premise only if the premise begins with a universal quantifier having the entire rest of the sentence in its scope), you must first remove “(x)”, and then remove “(y)”:

3. (y)(Fay É Cya)                    1 UI

4. Fab É Cba                           2 UI

You can now complete the proof using familiar rules:

5. Cba                                      2, 4 MP

There are restrictions on the use of UI that you must keep in mind, especially when working on proofs containing relational predicates:

RESTRICTION A: When using UI, you must use the same constant, or the same quasivariable, to replace multiple instances of a given bound variable.

For example, the following is invalid:

1. (x)(y)(Fxy É Cyx)               p

2. Fab                                      p          /\ Cba

3. (y)(Fay É Cyb)                    1 UI     invalid

Both occurrences of “x” in 1 must be replaced with the same constant or quasivariable. [This is true not only for UI but for EI as well. However, when applying EI, you can never use individual constants to replace bound variables... you must replace bound variables with quasivariables.]

RESTRICTION B: When using UI to replace a bound variable with a quasivariable, the quasivariable you introduce must be a genuine quasivariable and therefore free, i.e., it cannot be bound to any remaining quantifiers (if it were bound, it would not actually be a quasivariable!).

So the following is invalid:

1.      (x)(y)(Fxy É Cyx)             p

2.      (y)(Fyy É Cyy)                  1 UI     invalid

This is invalid because the “y”s in line 2 that replace the “x”s in line 1 are not free and are thus not really quasivariables; they are bound to the “(y)” in line 2 and are thus bound variables.

[8.5.2.] Universal Generalization.

If the conclusion of your proof is a universal quantification, then you will need to put the quantifiers back on at the end of the proof. For example:

Pxy= x is a parent of y

1. (x)(y)(Fxy É Cyx)               p

2. (x)(y)(Cyx É Pxy)               p          /\ (x)(y)(Fxy É Pxy)

Here you would first apply UI as before (except that you will need to instantiate the first two premises using quasivariables rather than constants; otherwise, at the end of the proof you could not use UG to reintroduce the universal quantifiers).

3. (y)(Fzy É Cyz)                    1 UI     [here “z” is a quasivariable]

4. Fzw É Cwz                          3 UI     [here “w” is a quasivariable]

5. (y)(Cyz É Pzy)                    2 UI

6. Cwx É Pzw                          5 UI

Now apply HS to move closer to the desired conclusion...

7. Fzw É Pzw                          4, 6 HS

Now you’ll need to add the quantifiers back using UG. You must add them back in reverse order (starting from the inside and then moving outwards), since UG can only be applied to entire lines:

8. (y)(Fzy É Pzy)                     7 UI     [replaces quasivariable “w” with bound variable “y”]

9. (x)(y)(Fxy É Pxy)                8 UI     [replaces quasivariable “z” with bound variable “x”]

Reminder: there are serious restrictions on UG:

A.     When using UG, you cannot replace an individual constant with a bound variable; you can only replace a quasivariable with a bound variable

B.     When using UG, the quasivariable you are replacing cannot have been introduced into the proof with EI.

C.     When using UG, the quasivariable you are replacing cannot occur free in an undischarged assumed premise.

Here is a further restriction that is especially important in relational predicate logic proofs:

RESTRICTION C: When using UG, you must replace every occurrence of the quasivariable you are replacing.

As an illustration of this restriction, consider that the following move would be invalid:

7. Fzw É Pzw                          4, 6 HS

8. (y)(Fzy É Pzw)                    7 UG    invalid

This move is invalid because it has replaced only one of the “w”s in 7 with a bound individual variable. In this example, if you replace one “w” (loosely speaking, if you “bind” one “w” quasivariable to the universal quantifier you’re introducing), you must bind the other.

[There is no analogous restriction on EG, as we will see below.]

[8.5.3.] Existential Instantiation.

1. (\$x)(\$y)Fxy                         p

2. (x)(y)(Fxy É Cyx)               p          /\ (\$x)(\$y)Cyx

As with UI, you should use EI multiple times to remove multiple quantifiers, working from the outside in.

Remember the original restrictions on EI:

A.     When using EI, the bound variable you are replacing must be replaced with a quasivariable rather than an individual constant.

B.     When using EI, the quasivariable you are introducing cannot already occur on any other line.

And remember, because of that second restriction, when you have a choice between applying UI first and applying EI first, you should always apply EI first.

3. (\$y)Fxy                                1 EI                  [“x” is a quasivariable]

4. Fxy                                      3 EI                  [“y” is a quasivariable]

5. (y)(Fxy É Cyx)                    2 UI                 [“x” is a quasivariable]

6. Fxy É Cyx                           5 UI                 [“y” is a quasivariable]

7. Cyx                                      4, 6 MP

And as with UG, you should use EG multiple times to add multiple quantifiers, working from the inside out (we will look more closely at EG below):

8. (\$y)Cyx                               7 EG

9. (\$x)(\$y)Cyx                         8 EG

Some further restrictions:

RESTRICTION D [analogous to restriction A for UI]: when you use EI to remove an existential quantifier, you must replace every occurrence of the variable you are freeing with the same quasivariable; i.e., you cannot do the following:

1. (\$x)(\$y)(Fxy · Cyx)                        p

2. (\$y)(Fzy · Cyw)                  1 EI      invalid

[8.5.4.] Existential Generalization.

Unlike with UG, when you use EG to add an existential quantifier, you do not have to bind every occurrence of a given quasivariable (or constant).

In other words, when you apply EG to bind a constant or quasivariable to a quantifier, you are permitted to bind only some occurrence of that constant or quasivariable in the sentence and to leave other occurrences unbound.

Nx = x is a narcissist

Lxy = x loves y

1. (x)(Nx É Lxx)                     p

2. Na                                        p          /\ (\$x)Lax

3. Na É Laa                             1 UI

4. Laa                                      2, 3 MP

5. (\$x)Lax                                4 EG

From the premises that all narcissists love themselves and Adam is a narcissist, it follows that Adam loves someone (namely, himself), and that is what the conclusion says.

However, there is one restriction on EG:

RESTRICTION E: When you use EG to replace an individual constant or quasivariable with a bound variable, there must be no additional occurrences of the variable to be bound contained in the expression to which you are applying EG. (This restriction is captured by the statement of EG on the inside cover of your textbook and on p.245.)

For example, the following move is invalid:

1. (\$x)(\$y)Lxy             p

2. (\$y)Lxy                   1 EI

3. Lxy                          2 EI

4. (\$x)Lxx                   3 EG    invalid -- there is an “x” in 3, so you cannot replace the quasivariable “y” in 3 with a bound “x.”

Without this restriction, the following obviously invalid argument would be valid: There is someone who loves someone (or other). Therefore, there is someone who loves himself.

[8.5.5.] Premises with Mixed Quantifiers.

Example #1 (p.242)

1. (\$x)(y)(Fx · Fxy)                 p

Because we have to work from the outside in when removing multiple quantifiers, we have to apply EI first:

2. (y)(Fx · Fxy)                                   1 EI

3. Fx · Fxy                              2 UI

Example #2

1. (x)(\$y)(Fx É Gxy)               p

Because we have to work from the outside in when removing multiple quantifiers, we have to apply UI first:

2. (\$y)(Fx É Gxy)                    1 UI     [“x” is a quasivariable]

Since “x” is free in line 2, you cannot use “x” to instantiate the bound variable “y” you will eliminate when you apply EI. In other words, you cannot make the following move...

3. Fx É Gxx                             2 EI      invalid

since quasivariable “x” occurs in an existing line in the proof. A correct application of EI to line 2 would be:

3. Fx É Gxy                             2 EI

SUMMARY OF THE NEW RESTRICTIONS:

RESTRICTION A: When using UI, you must use the same constant or quasivariable to replace every instance of the same bound variable:

1. (x)(y)(Fxy É Cyx)               p

2. Fab                                      p

3. (y)(Fay É Cyb)                    1 UI     invalid

RESTRICTION B: When using UI, any quasivariable you introduce must be a genuine quasivariable, not bound to any remaining quantifiers.

1.      (x)(y)(Fxy É Cyx)             p

2.      (y)(Fyy É Cyy)                  1 UI     invalid

RESTRICTION C: When using UG to replace a quasivariable with a bound individual constant, you must replace every occurrence of that quasivariable with the same variable.

1. Fzw É Pzw                          p

2. (y)(Fzy É Pzw)                    1 UG    invalid

RESTRICTION D: When using EI, you must replace every occurrence of the relevant individual variable with the same quasivariable.

1. (\$x)(\$y)(Fxy · Cyx)                        p

2. (\$y)(Fzy · Cyw)                  1 EI      invalid

RESTRICTION E: When using EG to replace an individual constant or quasivariable with a bound variable, there must be no additional occurrences of the variable to be bound contained in the expression to which you are applying EG.

1. Lxy                          p

2. (\$x)Lxx                   2 EG    invalid there is an “x” in 3, so you cannot replace the quasivariable “y” in 3 with a bound “x.”

[8.5.6.] Strategy. (Ch.10:8)

Chapter 10:8 contains the following strategic tips.

1.      Some premises will contain both a universal quantifier and an existential quantifier, with the latter falling within the scope of the former. For example,

1. (x)(\$y)~Gxy                        p

In such cases, you are forced to remove the universal quantifier before removing the existential quantifier. Your strategy should still be to remove the existential quantifier at the earliest possible opportunity. For example,

1. (x)(Mx É Px)                       p

2. (x)(\$y)~Gxy                        p

3. (\$y)~Gxy                 2 UI

4. ~Gxy                                   3 EI

You should aim to remove the (\$y) in line (2) as early in the proof as possible. This means you will need to take off both quantifiers in (2) before dealing with the quantifier in (1).

Notice that you cannot move from (3) to the following...

4. ~Gxx                                   2 EI      invalid

...because quasivariable “x” is already in (3).

2.      In order to complete some proofs, you must use new letters for quasivariables when applying EI and UI. For example (from p.244):

1. (\$x)(y)Fxy                           p

2. (y)(x)(Fyx É Gxy)               p

3. (y)Fxy                                  1 EI

4. Fxy                                      3 UI

5. (x)(Fxx É Gxx)                    2 UI     invalid

The move from 2 to 5 is invalid, since UI requires that you replace a bound variable with either an individual constant or a quasivariable. In 5, the “x” that is introduced to replace the individual variable “y” in line 2 is bound to the universal quantifier, so it is not really a quasivariable. We can avoid this problem by using letters other than “x” and “y” for the quasivariables introduced at lines 3 and 4:

1. (\$x)(y)Fxy                           p

2. (y)(x)(Fyx É Gxy)               p

3. (y)Fwy                                 1 EI

4. Fwz                                      3 UI

5. (x)(Fwx É Gxw)                  2 UI

6. Fwz É Gzw                          5 UI

3.      Some proofs can be solved only with IP (Indirect Proof).

An example of such a proof, from p.249:

1. (x)(\$y)(Fx · Gy)      p  /\ (x)Fx

One way to proceed is simply to remove the quantifiers, as we have done in the past:

2. (\$y)(Fx · Gy)           1 UI

3. Fx · Gy                    2 EI  [notice that “x” is free in this line, which is justified by EI]

4. Fx                            3 Simp

5. (x)Fx                                    4 UG    invalid

The move from 4 to 5 is invalid, because of the restriction on UG that says that UG cannot be used to bind a quasivariable that appears free in a line justified by EI. Even though the quasivariable in question was not introduced into the proof by EI, it still appears in a line justified by EI, and so cannot be replaced with a bound variable using UG.

The way around this impasse is to open an indirect proof at the very start:

1. (x)(\$y)(Fx · Gy)      p  /\ (x)Fx

2. ~(x)Fx                      AP / \ (x)Fx

3. (\$x)~Fx                    2 QN

4. ~Fx                          3 EI

5. (\$y)(Fx · Gy)           1 UI

6. Fx · Gy                    5 EI

7. Fx                            6 Simp

8. Fx · ~Fx                  4, 7 Conj

9. (x)Fx                                    2-8 IP

Exercise 10-9 (pp.246-47)

·         identify the mistakes in the proofs (each of them has at least one).

·         do all of these for next time

Exercise 10-10 (p.249)

·         complete all of these for next time

OPTIONAL: Exercise 10-11 (pp.250)

·         some of these are very difficult!

·         we won’t necessarily cover these in class, but I will email everyone the answers after class

Stopping point for Thursday April 12. For next time

·         complete ex.10-6, 10-7, 10-9 and 10-10; 10-11 is optional, for additional practice;

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