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6.2: Some Derived Rules

Problem 5-7(q) posed a special difficulty:

(Vx)~Fx                          1 | (Vx)~Fx                    P
~(∃x)Fx                          2 |       | (∃x)Fx              A
                                      3 |       |   a| Fa              A
                                      4 |       |     | (Vx)~Fx     1, R
                                     5 1    2|    3| ~Fa           4, VE  
                                             ?       ?    ?

We would like to apply ~I to derive ~(∃x)Fx. To do this, we need to get a contradiction in subderivation 2. But we can use the assumption of subderivation 2 only by using ∃E, which requires starting subderivation 3, which uses 'a' as an isolated name. We do get a sentence and its negation in subderivation 3, but these sentences use the isolated name 'a', so that we are not allowed to use ∃E to bring them out into subderivation 2 where we need the contradiction. What can we do?

   We break this impasse by using the fact_that from a contradiction you can prove anything. Be sure you understand this general fact before we apply it to resolving our special problem. Suppose that in a derivation you have already derived X and ~X. Let Y be any sentence you like. You can then derive Y:

 | .
 | .
 | .
 | X
 | ~X
 |      |~Y    A
 |      | X     R
 |      | ~X   R
 | Y             RD

   We can use this general fact to resolve our difficulty in the following way. Since anything follows from the contradiction of 'Pa' and '~Pa', we can use this contradiction to derive a new contradiction, 'A & ~A', which does not use the name 'a'. ∃E then licenses us to write 'A & ~A' in derivation 2 where we need the contradiction.

   To streamline our work, we will introduce several new derived rules. The first is the one I have just proved, that any sentence, Y, follows from a contradiction:

Contradiction

| X
| ~X
| Y      CD

In practice, I will always use a standard contradiction, 'A & ~A', for Y. I will also use a trivial reformulation of the rules ~I and Rd expressed in terms of a conjunction of a sentence and its negation where up to now these rules have, strictly speaking, been expressed only in terms of a sentence and the negation of the sentence on separate lines:

Negation Introduction                                                     Reductio
|    | X                                                                           |    | ~X           
|    | Y & ~Y                                                                   |    | Y & ~Y
|   ~X     ~|                                                                    | X                 RD

   These derived rules enable us to deal efficiently with problem 5-7(q) and ones like it:

(Vx)~Fx                          1 | (Vx)~Fx                P
~(∃x)Fx                          2 |     | (∃x)Fx            A
                                      3 |     |  a| Fa             A
                                      4 |     |    | (Vx)~Fx    1, R 
                                      5 |     |    | ~Fa           4, VE
                                      6 |     |    | A & ~A      3, 4, CD
                                      7 |     | A & ~A            2, 3-6 ∃E
                                      8 | ~(∃x)Fx                 2-7, ~I

   Let's turn now to four more derived rules, ones which express the rules of logical equivalence, ~V and ~∃, which we discussed in chapter 3. There we proved that they are correct rules of logical equivalence. Formulated as derived rules, you have really done the work of proving them in problems 54q) and (r) and 5-7(q) and (r). To prove these rules, all you need do is to copy the derivations you provided for those problems, using an arbitrary open sentence (. . . u . . .), with the free variable u, instead of the special case with the open sentence 'Px' or 'Fx' with the free variable 'x'.

Negated Quantifier Rules

| ~(Vu)(. . . u . . . )                                   | (∃u)~(. . . u . . .)
| (∃u)~(. . . u . . .) ~V                               | ~(Vu)(. . . u . . . ) ∃~       

| ~(Vu)(. . . u . . . )                                   | (Vu)~(. . . u . . .)
| (∃u)~(. . . u . . .) ~∃                               | ~(∃u)(. . . u . . . ) V~       

   A word of caution in using these negated quantifier rules: Students often rush to apply them whenever they see the opportunity. In many cases you may more easily see how to get a correct derivation by using these rules than if you try to make do without the rules. But often, if you work hard and are ingenious, you can produce more elegant derivations without using the quantifier negation rules. In the following exercises, use the rules so that you have some practice with them. But in later exercises, ' be on the lookout for clever ways to produce derivations without the quantifier negation rules. Instructors who are particularly keen on their students learning to do derivations ingeniously may require you to do later problems without the quantifier negation rules. (These comments do not apply to the derived contradiction rule and derived forms of ~I and RD rules. These rules iust save work which is invariably boring, so you should use them whenever they will shorten your derivations.)

Exercise

6-2.

a) (Vx)Px                             b)  (Vx)(Fx ⊃ G)                 c) ~(Vx)(Vy)Lxy
    (Vx)~Qx                              (Vx)(Gx ⊃ Hx)                   (∃x)(∃y)~Lxy
    ~(∃x)(Px = Qx)                       ~(∃x)HX               
                                                  ~(∃x)Fx

d) ~(∃x)(∃y)Lxy                    e)     ~(∃x)(Px v Qx)           f) ~(Vx)(Px & Qx)      
     (Vx)(Vy)~Lxy                        (Vx)~PX & (Vx)~Qx           (∃x)~Px v (∃x)~Qx

g) (Vx)[~(∃y)Rxy & ~(∃y)Ryx]           h) (∃x)[Px ⊃ (Vy)(Py ⊃ Qy)]
            (Vx)(Vy)~Rxy                                   ~(∃x)QX                
                                                                   ~(Vx)Px

i) (∃y)(∃z)[(Vx)~Rxy v (Vx)~Rxz]
     ~(Vy)(Vz)(∃x)(Rxy & Rxz)

 

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