Practice Midterm 1 (A)

These are solutions to midterm #2.

Keep in mind that for the definitions and short answer problems, answers other than those given might receive full credit if they are alos sifficiently good. And also there are many ways that each proof can be done. Only one is shown below.

Give a one sentence definition of the following terms.

1. Assumption

The first line in a subproof (either conditional or indirect).

2. Nested subproof

A subproof that is within another subproof.

3. Accessible line

Any line that is not within a subproof that has been closed at the surrent point in the proof.

4. Main operator

The operator that determines the logical form of the statement as a whole. (Or: determines the truth value of the statement.)

5. Explicit contradiction

Any statement conjoined to its own negation.

Part Two: Short Answer. (2 problems x 6 points each = 12 points)

6. Explain the difference(s) between a rule of inference and a replacement rule.

There are three differences. First, Inference rules must be applied only to entire lines, whereas replacement rules can be applied to parts of lines as well as entire lines. Second, inference rules only work in one direction, whereas replacement rules go both ways. The third difference, which explains the first two, is that replacement rules embody implications, whereas replacement rules embody equivalences.

7. Explain the principle behind indirect proof.

In indirect proof one proves a statement by showing that the negation of that statement must be false. This is done in a subproof, in which the negation of the statement is assumed, and an explicit contradiction derived. Since a contradiction must be false, and since the rules are truth preserving, some line on which the rules were applied must be false, and since the only line whose truth is in question is the assumption, it must be false. Hence its negation must be true.

Part Three: Proofs. (9 proofs x 7 points each = 63 points)

8. (use any inference or replacement rules, but not CP or IP)

1. ~(~A v B) v (P v ~C)
2. ~P
3. A -> P / .: ~C

4. ~A 2, 3 MT
5. ~A v B 4 Add
6. P v ~C 1, 5 DS
7. ~C 2, 6 DS

9. (use any inference or replacement rules, but not CP or IP)
01. ~C -> ~B
02. A
03. A -> B / .: C

4. B 2, 3 MP
5. C 4, 1 MT

10. (use any inference or replacement rules, but not CP or IP)
1. ~(A v L)
2. P v Q
3. Q -> (R & A)
4. P -> (K v L) / .: K

5. ~A & ~L 1 DeM
6. ~L 5 Simp
7. ~A 5 simp
8. ~R v ~A 7 add
9. ~(R & A) 8 DeM
10. ~Q 9, 3 MT
11. P 2, 10 MT
12. K v L 4, 11 MP
13. K 12, 6 DS

11. (use any inference or replacement rules, and you must use IP)
1. (N v S) (L & ~M)
2. ~L v M / .: ~N

13. (use any rules or methods)
1. B -> (K & M)
2. (B & M) (P <-> ~P) / .: ~B

03. |-> B AIP
04. | K & M 3, 1 MP
05. | M 4 Simp
06. | B & M 3, 5 Conj
07. | P <-> ~P 2, 6 MP
08. | (P -> ~P) & (~P -> P) 7 BE
09. | P -> ~P 8 simp
10. | ~P -> P 8 Simp
11. | ~P v ~P 9 CE
12. | ~P 11 Dup
13. | P v P 10 CE
14. | P 13 Dup
15. |_ P & ~P 14, 12 conj
16. ~B 3-15 IP

14. (use any rules or methods)
1. Z & ~Q
2. K -> {W -> (~Z & O)} / .:K -> (~W & ~Q)

15. (use any rules or methods)
/ .: {Z -> (X v Y)} -> {~X -> (~Y -> ~Z)}

01. |-> Z -> (X v Y) ACP
02. | |-> ~X ACP
03. | | |-> ~Y ACP
04. | | | ~X & ~Y 2, 3 conj
05. | | | ~(X v Y) 4, DeM
06. | | |_ ~Z 1, 5 MT
07. | |_ ~Y -> ~Z 3-6 CP
08. |_ ~X -> (~Y -> ~Z) 2-7 CP
09. {Z -> (X v Y)} -> {~X -> (~Y -> ~Z)} 1-8 CP

16. (use any rules or methods)
/ .: (P -> Q) v (Q -> P)

01. |-> ~{(P -> Q) v (Q -> P)} AIP
02. | ~(P -> Q) · ~(Q -> P) 1 DeM
03. | ~(P -> Q) 2 simp
04. | ~(Q -> P) 2 simp
05. | ~(~P v Q) 3 CE
06. | ~(~Q v P) 4 CE
07. | P & ~Q 5 DeM
08. | Q & ~P 5 DeM
09. | Q 8 simp
10. | ~Q 7 simp
11. |_ Q & ~Q 9, 10 conj
12. (P -> Q) v (Q -> P) 1-11 IP