[9.5.] 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.

 

 

[9.5.1.] 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 both quantifiers from a given premise. You will apply it one time to remove the first quantifier (and 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: You cannot use two different constants, or two different quasivariables, to replace multiple instances of the same 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, except that when applying EI, you can never use individual constants to replace bound variables... you must replace bound variables with quasivariables.]

 

 

RESTRICTION B: When replacing a bound variable with a quasivariable by way of UI, 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.

 

 

[9.5.2.] 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. Cwz 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 “w” with a bound “y”]

9. (x)(y)(Fxy É Pxy)                 8 UI     [replaces “z” with a bound “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 binding cannot occur free in a line justified by EI.

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

 

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

 

RESTRICTION C: You must bind every occurrence of a given quasivariable or constant when applying UG. In other words, when you apply UG to bind a constant or quasivariable to a quantifier, you must bind every occurrence of that constant or quasivariable in the sentence to which you are applying UG.

 

So in the proof given above, 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 bound only one of the “w”s in 7 to the new quantifier introduced in 8. In this example, if you bind one “w” to a universal quantifier, you must bind the other.

 

[The same is not true for EG, as we will see below.]

 

 

[9.5.3.] 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: As with 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

 

 

 

[9.5.4.] 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 quasivariabl 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

a = Andrew

 

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 Andrew is a narcissist, it follows that Andrew 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.241.)

 

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.

 

 

[9.5.5.] Premises with Mixed Quantifiers.

 

Example #1 (p.238)

 

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

 

3. Fx É Gxy                              2 EI

 

Exercise 10-9 (pp.243-44)

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

·         do all of these for next time

 

 

Stopping point for Wednesday April 16. For next time, complete ex.10-9, and read ch.9-8 (pp.244-45).