Formal Logic 101 – Part 13: Replacement Rules (Part 4)

Welcome back, welcome back! Glad to have you. Come on in, take a seat, kick your shoes off, and get ready for yet another post in our series in formal logic. We only have three more replacement rules. We’re only going to get to one in this post, because we also have to learn our last operator “If and only if”, so that’ll be the post for today, and we’ll cover the last two next time. 

Biconditional Exchange (BE)

In our last post, we discussed Conditional Exchange. In this case we will be discussing Biconditional Exchange. Do not let the name fool you, the two are not really all that similar. We’ve previously discussed four operators “If/Then”, “And”, “Or”, and “Not”. To understand Biconditional Exchange, we need to add one more operator “If and Only If”.

To start, let’s briefly recap “If/Then” so we can contrast it with “Only If”, and then see how the two behave together.

Do note that the “If” does not necessarily have to be at the beginning of the sentence. For instance, “If she Wants to dance, then she’ll Dance”, and “She’ll Dance if she Wants to” are both symbolized W→D (where “W” is “she Wants to dance”, and “D” is “she’ll Dance”). In both cases “she’ll dance” is dependent on whether “she wants to”. Since “she’ll dance” is the dependent part, it goes in the back.

Now consider the phrase:
“You’ll make it into Harvard only if you complete the Application”.

Let “H” be “you’ll make it into Harvard”, and let “A” be “you complete the Application”. From what we just said about “If/Then” it might be tempting to symbolize it as A→H, but fight back against that temptation. Ask yourself which of the following statements means the same thing as our original phrase:

1. “If you complete the application, then you’ll be let into Harvard”
Or
2. “If you want to be let into Harvard, then you must complete the application”

As a way of helping to keep it straight, I recommend looking to the real-world as our example: After all, it’s clear that not everyone who applies to Harvard is let in, but I’m not aware of anyone who was let into Harvard without even applying.

Keeping that in mind, if you thought it was the second option: Congrats! Add a gold star to the smart noodle board. (And if you thought it was #1. Add a gold star to the “noodle that learned something new today” board.) The Correct symbolization of “You’ll make it into Harvard only if you complete the Application” is “H→A”.

Now, say that Harvard was falling on hard times and for some reason just couldn’t get any new students. Suppose they loosened up the restrictions a bit and said “You will be let into Harvard, only if you complete the application… but if you complete the application, you will be admitted into Harvard.”

We might symbolize that as “(H→A) ^ (A→H)” and that would be correct. However, you can also symbolize it by using our last operator, “Biconditional” (or “If and only if) and symbolize it as “H↔A”, or “You will be let into Harvard, if and only if you complete the application”.

Biconditional Exchange simply states that “(H→A) ^ (A→H)” is interchangeable with “H↔A”.


Biconditional is the last operator that we'll learn in this course, So we now have all of our operators, all of our inference rules, and eight of the the ten replacement rules. We'll cover those last two replacement rules next time. In the meantime, practice problems have been included below.


Practice

Inference Rules:

Replacement Rules:
Argument A
1. AvB→(A^B)        P – Prove A↔B

Argument B
1. C→(D→~C)        P
2. C↔D                    P – Prove ~C^~D

Argument C
1. B→[(F^G)↔D]        P
2. (C→D)→~A            P
3. ~F→(D^E)               P
4. ~G→~B                   P – Prove ~(A^B)

Answers

As always these are not necessarily the only possible answers.

Argument A
1. AvB→(A^B)               P – Prove A↔B
2. A                                 A – Prove A→B
3. AvB                            2 add
4. A^B                            1,3 MP
5. B                                 4 simp
6. A→B                          2-5 CP
7. B                                 A – Prove B→A
8. AvB                            7 add *
9. A^B                            1,8 MP
10. A                               9 simp
11. B→A                        7-10 CP
12. (A→B)^(B→A)        6,11 Conj
13. A↔B                        12 BE

*Remember from Part 9, once you use CP or IP the assumption and any lines you reached by using it are off limits. Line 6 tells me that I cannot use 2,3,4, or 5 ever again (line 6 itself, however, is fair game). That’s why I need to make line 8 “AvB” rather than reusing line 3.


Argument B
1. C→(D→~C)            P
2. C↔D                        P – Prove ~C^~D
3. (C→D)^(D→C)       2 BE
4. C                              A – Prove contradiction
5. C→D                       3 simp
6. D                              4,5 MP
7. D→~C                     1,4 MP
8. ~C                            6,7 MP
9. C^~C                       4,8 conj
10. ~C                          4-9 IP
11. D                             A – Prove contradiction
12. D→C                      3 simp
13. C                             11,12 MP
14. D→~C                    1,13 MP
15. ~C                          11,14 MP
16. C^~C                      13,15 conj
17. ~D                          11-16 IP
18. ~C^~D                   10,17 Conj

Argument C
1. B→[(F^G)↔D]                        P
2. (C→D)→~A                           P
3. ~F→(D^E)                              P
4. ~G→~B                                  P – Prove ~(A^B)
5. A^B                                         A – Prove contradiction
6. B                                             5 simp
7. (F^G) ↔D                              1,6 MP
8. [(F^G)→D]^[D→(F^G)]         7 BE
9. A                                            5 simp
10. ~~A                                     9 DN
11. ~(C→D)                              2,10 MT
12. ~(~CvD)                             11 CE
13. ~~C^~D                             12 DeM
14. (F^G)→D                            8 simp
15. ~D                                       13 simp
16. ~(F^G)                                14, 15 MT
17. ~Fv~G                                16 DeM
18. ~~B                                     6 DN
19. ~~G                                     4,18 MT
20. ~F                                        17,19 DS
21. D^E                                      3,20 MP
22. D                                          21 simp
23. D^~D                                  15,22 conj
24. ~(A^B)                                5-23 IP

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