construction-categorical-derivation-and-construction-derivation-valid-argument
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practice_assignment_7.docPractice Assignment 7
Here are a few sample questions.
a) An ‘or’ as a premise
W v ~T
(W & G) > Q
G & L
~T = (Q & B)
Q & L
b) Premise used as half of a contradictory pair
A = D
~(T > D)
~A
c) Using a contradictory pair
F > ~D
R & F
(Y & D) > E
d) v in conclusion
A = H
~B > ~G
A v ~B
H v ~G
Answers
a) It’s important to have a game plan before you start a derivation. First, look at the conclusion. The major operator will often tell you what rule you need to start off with. Here, the conclusion has an &. We can easily get an L out of 3 so we have to concentrate on getting a Q.
1. W v ~T Pr.
2. (W & G) > Q Pr.
3. G & L Pr.
4. ~T = (Q & B) Pr.
Next, take a look at each of the premises and see what rules you expect to use. Line 1 has an v so we expect to use an vE. That’s a tough rule so we should do it first. The rule says that we need two sub-derivations, one assumes the W, the other assumes the ~T. We have to prove the same thing in both but it can be anything we want. Since we need a Q, let’s try to prove a Q in each.
5. | W A.
|-----
|
|
|
| Q
|~T A.
|-----
|
|
|
|Q
Okay, let’s start with the first sub-derivation. We need a Q. Take a look at the premises. There is a Q in 2 that we can get out but we’d need to have a G (we can get that from 3).
5. | W A.
|-----
6. |(W & G) > Q reit 2
7. |G & L reit 3
8. |G &E 7
9. |W & G &I 5, 8
10. |Q >E 6, 9
Now we need to work on the second sub-derivation.
11. |~T A.
|-----
12. |~T = (Q & B) reit 4
13. |Q & B =E 11, 12
14. |Q &E 13
15. Q vE 1, 5 – 10, 11 - 14
16. L &E 3
17. Q & L &I 16, 17
b)
Again, look first at the conclusion. The major operator is a ~ so we expect to do an ~I. Take a look at the premises. We could break up the first premise as long as we have one side of the =. What about the second premise? I don’t have a rule that will allow me to break up the ~ of a >. The only thing that this premise can be used for is as one half of a contradictory pair. The other half will be T>D.
1. A = D Pr.
2. ~(T > D) Pr.
| A A.
|-----
|
|
| T > D We have to try to prove this.
|~(T > D) reit. 2
~A
Now, we have to get T > D. The major operator is the > so we set up a >I.
1. A = D Pr.
2. ~(T > D) Pr.
3. | A A.
|-----
4. | | T A.
| |----
| |
| |
| | D
| T > D >I
|~(T > D) reit. 2
~A
Now we just have to prove D from what is above.
1. A = D Pr.
2. ~(T > D) Pr.
3. | A A.
|-----
4. | | T A.
| |----
5. | |A = D reit 1
6. | |A reit 3
7. | | D =E 5, 6
8. | T > D >I 4 - 7
9. |~(T > D) reit. 2
10.~A ~I 3, 8, 9
c)
The major operator in the conclusion is a > so we should start off by setting up a >I
1. F > ~D Pr.
2. R & F Pr.
3. | Y & D A.
|----------
|
|
|
| E
(Y & D) > E > I
Now, we have to prove E. The problem is there aren’t an Es in the premises. Hmmm. Well let’s see what else we can do with the premises. I have an F in 2 which can be used with 1 to get a ~D. Does this help? I have a D in 3 so it looks like I have a contradictory pair at my disposal. Ahhh, I get that warm and fuzzy feeling like everything is going to be okay. Once I have a contradictory pair, I can prove anything that I want through an indirect proof. Set up an indirect proof (~E + ~I) for E.
1. F > ~D Pr.
2. R & F Pr.
3. | Y & D A.
|----------
4. | |~E A.
| |----
| |
| |
| |
|~~E ~I
| E ~E
(Y & D) > E > I
Now, the finish off the ~I we need a contradictory pair, and we have one.
1. F > ~D Pr.
2. R & F Pr.
3. | Y & D A.
|----------
4. | |~E A.
| |----
5. | |R & F reit 2
6. | |F &E 5
7. | |F > ~D reit 1
8. | |~D >E 6, 7
9. | | Y & D reit 3
10. | |D &E 9
11. |~~E ~I 4, 8, 10
12. | E ~E 11
13. (Y & D) > E > I 3 - 12
d)
The third premise has an v, so as we saw above, we expect to use an vE. Set that up first.
1. A = H Pr.
2. ~B > ~G Pr.
3. A v ~B Pr.
4. |A A.
|---
|
|
|
|~B A
|-----
|
|
|
The question now is, what do we try to prove in each sub-derivation. The rule tells us that it must be the same but we can choose whatever we want. Well, the conclusion is H v ~G so why don’t we try to prove H, then we can or the ~G later.
4. |A A.
|---
5. |A = H reit 1
6. |H =E 4, 5
7. |~B A
|-----
8. |~B > ~G reit 2
|~G >E 7, 8
|
|H
The first sub-derivation was easy enough but now we have run into a problem with the second. How do we get from ~ G to H? It seems impossible.
The solution lies in keeping the information in the conclusion together as long as possible. Instead of trying to get an H in each sub-derivation, try to get H v ~G
4. |A A.
|---
5. |A = H reit 1
6. |H =E 4, 5
7. |H v ~G vI 6
8. |~B A
|-----
9. |~B > ~G reit 2
10. |~G >E 8, 9
11. |H v ~G vI 10
12. H v ~G vE 3, 4 – 7, 8 – 11
The advantage with keeping the v together is that there are two ways to make an v. Each way was used in the proof.
Here’s what we’ve learned so far .....
1. Take a look at the major operator in the conclusion. That will tell you what rule you will have to use to get the conclusion. If the conclusion is a single letter then see if you can get the conclusion simply out of the premises. If neither of these options work then try an indirect proof.
2. Take a look at the premises. What can you break up?
3. If you have an v in a premise then be prepared to do an vE. Set that up first.
4. If you have a premise that cannot be broken up (such as ~(A>B ), ~(A=B), ~(A&B), ~(AvB)) then use it as half of a contradictory pair in ~I.
5. If when you break up the premises you find you have a contradictory pair at your disposal then you can use it in an indirect proof to prove anything you want. Always look for a contradictory pair in the premises.
6. If you have an v in the conclusion you may want to keep it together as long as you can. There are two ways to make an v. Either find the left part and v the right, or find the right and v the left. It may be helpful to leave these options open.
These derivations may seem tough at first but will get much easier as you do more of them. If you get stuck, take a break and come back to a question later. The fresh perspective might help. Always look at your major operators, they tell you what rules to use.
