Formal Logic 101 – Part 12: Replacement Rules (Part 3)

As promised last time, today we’ll be going over some of the most commonly used replacement rules. 

DeMorgan’s Theorem

Demorgan’s is handy partially because it helps you get propositions out of parentheses, and also because it lets you change “And” statements into “Or” statements and vice versa. To illustrate this, let’s look at the following example:

Suppose I tell you that Kelly either plays Banjo or Ukulele. Letting “B” be “Kelly plays Banjo,” and “U” be “Kelly plays Ukulele,” we can symbolize my statement as: “BvU.”

Now suppose that you disagree with me, knowing that Kelly has never picked up an instrument in her whole life. “No,” you say, “It isn’t the case that Kelly plays either the Banjo or the Ukulele.” We can symbolize your statement as “~(BvU)”

Now, think about your claim that it isn’t the case that Kelly plays either Banjo or Ukulele “~(BvU).” 

Which of the following statements is equivalent to what you said:
1. Either Kelly doesn’t play banjo OR she doesn’t play Ukulele.
2. Kelly doesn’t play banjo AND she doesn’t play Ukulele.

If you answered #2. Congrats. Gold star for you.

DeMorgan’s Theorem can be used in 8 different ways (technically 16 ways since you have to proof that each way can go back and forth).

  1. ~(AvB) is equivalent to ~A^~B (and vice versa)
  2. ~(Av~B) is equivalent to ~A^B (and vice versa)
  3. ~(~AvB) is equivalent to A^~B (and vice versa)
  4. ~(~Av~B) is equivalent to A^B (and vice versa)
  5. ~(A^B) is equivalent to ~Av~B (and vice versa)
  6. ~(A^~B) is equivalent to ~AvB (and vice versa)
  7. ~(~A^B) is equivalent to Av~B (and vice versa)
  8. ~(~A^~B) is equivalent to AvB (and vice versa)

As such, proving DeMorgan's to be true requires 16 proofs that are each 7-10 lines long. In the interest of conserving space and time, I've posted a picture detailing the proofs below, so that interested readers can peruse them if they so desire.



Conditional Exchange

Conditional exchange is handy for changing transforming an “if/then” statement to an “Or” statement or vice versa. Conditional exchange establishes that “A→B” is the same as “~AvB”. (Alternately, “~A→B” is the same as “AvB”.

Think of Conditional exchange a lot like a disjunctive syllogism. We’ve been told “either A or B is true”. If we determine that A is false, then by process of elimination, it must be B. More formally, we can prove this as follows:

1. AvB             P – Prove ~A→B
2. ~A              Assumption
3. B                1,2 D.S.
4. ~A→B        2-3 C.P

And to prove that the reverse is true:

1. A→B              P – Prove ~AvB
2. ~(~AvB)        Assumption – Prove contradiction
3. ~~A^~B        1,2 DeMorgan’s
4. A^~B              3 D.N.
5. A                     4 Simp.
6. B                     1,5, M.P.
7. ~B                  4 Simp.
8. B ^ ~B            6,7, conj
9. ~~(~AvB)      2-8 I.P.
10. ~AvB            9 D.N.

So now we can change some combination of Conditional Exchange and Demorgan’s Theorem to transform “And”, “Or”, and “If/Then” statements into one another.

We have three more replacement rules to go, but I think this post has gone on quite long enough, so we’ll pick it up here next time. Fortunately, we’ve reached a point where we’ve covered enough replacement rules that I can fairly easily peruse my various books and course notes on logic and find examples that use replacement rules, but don’t use the last three that we haven’t covered, so there will be practice arguments below.


Practice

Inference Rules:

Replacement Rules:


Argument A
1. (AvB)→~(CvD)        P
2. (A^E)v~F                  P
3. F                                P – Prove ~C

Argument B
1. ~Av~(B^C)               P
2. ~[~Av(FvH)]            P
3. ~C→F                       P – Prove ~(BvH)

Argument C
1. (~Av~B)→~C          P
2. A→F                         P
3. (D^~E)→~F             P
4. ~(D→H)                   P
5. E→H                         P – Prove ~C

Argument D
1. (A→B)→~(C→D)     P
2. ~(AvF)                        P – Prove ~(DvF)



Answers


Argument A

1. (AvB)→~(CvD)           Premise (P)
2. (A^E)v~F                     P
3. F                                    P – Prove ~C
4. C                                   Assumption (A) – Prove contradiction
5. ~~F                              3 DN
6. A^E                               2,5 DS
7. A                                   6 simp
8. AvB                               7 add
9. ~(CvD)                         1,8 MP
10. ~C^~D                       9 DeM
11. ~C                              10 simp
12. C^~C                          4,11 conj
13. ~C                              4-12 IP

Argument B

1. ~Av~(B^C)                   P
2. ~[~Av(FvH)]                 P
3. ~C→F                           P – Prove ~(BvH)
4. ~~A^~(FvH)                2 DeM
5. ~~A                              4 simp
6. ~(B^C)                         1,5 DS
7. ~Bv~C                         6 DeM
8. ~(FvH)                         4 simp
9. ~F^~H                         8 DeM
10. ~F                              9 simp
11. ~~C                           3,10 MT
12. ~B                              7,11 DS
13. ~H                              9 simp
14. ~B^~H                       12,13 Conj
15. ~(BvH)                       14 DeM

Argument C

1. (~Av~B)→~C          P
2. A→F                          P
3. (D^~E)→~F              P
4. ~(D→H)                    P
5. E→H                          P – Prove ~C
6. ~~C                           A – Prove Contradiction
7. ~(~Av~B)                 1,6 MT
8. ~~A^~~B                  7 DeM
9. A^~~B                       8 DN
10. A                               9 simp
11. F                               2,10 MP
12. ~~F                          11 DN
13. ~(D^~E)                   3,12 MT
14. ~Dv~~E                   13 DeM
15. ~(~DvH)                  4 CE
16. ~~D^~H                  15 DeM
17. ~~D                          16 simp
18. ~~E                          14,17 DS
19. E                                18 DN
20. ~H                             16 simp
21. ~E                              5,20 MT
22. E^~E                          19,21 conj
23. ~C                              6-22 IP

Argument D

1. (A→B)→~(C→D)            P
2. ~(AvF)                               P – Prove ~(DvF)
3. DvF                                    A – Prove contradiction
4. ~A^~F                               2 DeM
5. ~A                                      4 simp
6. ~AvB                                 5 add
7. (~AvB)→~(C→D)           1 CE
8. ~(C→D)                            6,7 MP
9. ~(~CvD)                           8 CE
10. ~~C^~D                         9 DeM
11. ~D                                   10 simp
12. F                                      3,11 DS
13. ~F                                   4 simp
14. F^~F                               12,13 conj
15. ~(DvF)                            3-15 IP

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