LESSON #28

Rules of Replacement I

Reading Assignment: 7.3 (pp. 384-390)

Rules of replacement are slightly different from the previous rules. Instead of being mini proofs that let you derive something new, they only assert that if you have one of the statements listed, you can substitute the other, because they are logically equivalent. Because of this, you only need one line for justification.  Note that they work in both directions. You can start with the statement on either side and substitute the other. You can also use them on parts of lines, (i.e., you change one part of a line to something equivalent and just bring the rest of the line down with it). Since you don’t derive anything new with these rules, you couldn’t work a whole proof using just these rules. (except to prove something that was true by definition) You will use them to make using the first eight possible.

DeMorgan's

This is a very useful rule. Upon reflection you should be able to understand how to use it. ~(p · q) means not both p · q. Well, what does it mean to say not both? It means that either p is false or q is false or they are both false--anyway, p and q can't both be true at the same time. So ~(p · q) º ~p v ~q. 

On the other hand, ~(p v q) means it's not the case that either p or q.  In other words, they ate both not true.

 ~(p · q) º (~p v ~q) 

 

~(p v q) º ( ~ p · ~q)

Notice that to do DeMorgan's you have "~"s and "· "s and "v"s. (DeMorgan's does not work on É 's or º 's.) DeMorgan's involves a "distribution" of ~. The ~ moves from outside the parentheses to inside or vice versa. When it does this the connective must change from an "v" to an "· " or vice versa. 

DeMorgan's, then, allows you to do the following:

 If you have an · or an v inside parentheses with a ~ on the outside, you are allowed to get rid of the parentheses, change the connective (· à v; v à · ) and negate each side.

Commutativity--finally!

This is the rule many have been waiting for--that lets use the other disjunct in DS or the other conjunct in Simp. Commutativity simply allows you to flip the sides of an "v" and an "· ." This does not work for É or º

(p v q) º (q v p)

(p · q) º (q · p)

 

Associativity

 

Associativity lets you switch the placement of the parentheses when you have a statement of three components connected by either two "v"s or two "· "s. Notice that both signs must be the same. This will not work with a mixture of signs.

[p v (q v r)] º [(p v q) v r]

[p· (q · r)]  º [(p  · q)  · r]

Distribution

Distribution is useful but is certainly not used asfrequently as DeMorgan's. Distribution involves both· 's and v's in the same statement. There are three different components. One part is either inside the parenthesis with each of the other two or it is outside and separate from the parentheses. You can go back and forth between the two equivalences.

 [ p · (q v r)] º [(p · q) v (p· r)]

[p v (q · r) ]º [(p v q) · ( p v r)]

 

Notice that the main connective always changes between an · and an v. The other connective, then, is the one you put with the other components.

Double Negative

Double negative is quite simple. It works under the awareness that p º ~~p. e.g. "I ain't got none" literally means "I do have some." If you took what I said in the DeMorgan's section seriously, you are already doing double negation in your head, anyway.

p º ~~p

 

TIP: When working on part of a line instead of on the main connective, be sure not to forget to keep the rest of the line like it is and bring it down too.

CORRECT

e.g.  1. XXXXX
        2. XXXXX
        3. (~p · ~q) É (r · s)
        4. ~(p v q) É (r · s)        3 DM

 

WRONG!

e. g. 1. XXXXX
        2. XXXXX
        3. (~p · ~q) É (r · s)
        4. ~(p v q)                      3 DM

 

 

Logic Coach Assignment: 7.3 I all,  II 1-15, III 1-4

Assignment: (20 points each)

Using the first 13 rules of inference, derive the conclusions of the following symbolized arguments. Put the English argument into symbolic form first.

NOTE: Be sure to copy the problem neatly and correctly.

A) 1. ~E · ~F    
  2. ~(E v F) É ~G    
  3. ~G É H   // ~~H
         
         
B) 1. (N v M) v O    
  2. ~(M v O)    
  3. N É (P º Q)   // P º Q
         
         
C) 1. ~(F v G)    
  2. (~G v W) É (A · B)    
  3. T É ~B   // ~T
         
         
D) 1. (R ·P) v (M · N)    
  2. ~R   // M
 

    E) If women are by nature either passive or uncompetitive, then it is not the case that there are lawyers who are women. If men are by nature either insensitive or without the ability to nurture, then it is not the case that there are kindergarten teachers who are men. There are lawyers who are women and kindergarten teachers who are men. Therefore, it is not the case that either women by nature are uncompetitive or men by nature are without the ability to nurture. ( P, U, L, I, W, K)

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