NOTE: The format of the real first midterm will follow this format exactly. However, since I cannot duplicate our normal symbols for conjunction (dot), conditional (horseshoe) or biconditional (triple-bar) in HTML, I have used the following substitutions: conditional ->, biconditional <->, and conjunction &. The real exam will, however, use the normal symbols.
Midterm I
Philosophy 500 -- Introduction to Logic
Write your name here: ______________________________
The exam consists of 20 questions. Each is worth 5 points. Show all work.
Part One. Give a one sentence definition of the following terms.
1. Argument
A set of statements, one of which, the conclusion, is taken to be supported by the others, called premises.
2. Valid Argument
An argument such that if all the premises are true, the conclusion cannot be false.
3. Sentential Operator
An operator which takes one or more sentences as input and forms a single, more complex sentence as output.
4. Tautology
A sentence or statement that is always true.
5. Contingent sentence
A sentence or statement that can be either true or false.
Part Two. Short answer.
6. Explain the difference between a rule of inference and a replacement rule.
A rule of inference must be applied only to entire lines. Replacement rules can be applied to components of a statement within a line.
7. Explain the difference between a valid argument and a sound argument.
A sound argument is a valid argument that in fact has all true premises. A valid argument is one such that if the premises are true the conclusion cannot be false, but it may in fact have false premises and even a false conclusion.
Part Three. Truth tables.
In 9 and 10, use truth tables to determine if the given statement has a form that is contingent, a contradiction, or a tautology. You will first need to translate from the English to proper logical symbolism.
9. If I am covered with Cheez-Whiz, then if I am not covered with Cheez-Whiz, I like eggs.
C = I am covered with Cheez-Whiz.
E = I like Eggs.
C -> ( ~C -> I)
| C | I | ~C | ~C -> I | C -> (~C -> I) |
| T | T | F | T | T |
| T | F | F | T | T |
| F | T | T | T | T |
| F | F | T | F | T |
Since the statement has all T's in its column of the truth table, the statement is a tautology.
10. Both either sticky buns are inexpensive or I will go broke and I am not addicted to coffee, if and only if either sticky buns are inexpensive or both I will go broke and I am not addicted to coffee.
I = Sticky buns are inexpensive.
B = I will go broke.
C = I am addicted to coffee.
[(I v B) & ~C] <-> [I v (B & ~C)]
| I | B | C | ~C | I v B | (I v B) & ~C | B & ~C | I v (B & ~ C) | [(I v B) & ~C] <-> [I v (B & ~ C)] |
| T | T | T | F | T | F | F | T | F |
| T | T | F | T | T | T | T | T | T |
| T | F | T | F | T | F | F | T | T |
| T | F | F | T | T | T | F | T | T |
| F | T | T | F | T | F | F | F | T |
| F | T | F | T | T | T | T | T | T |
| F | F | T | F | F | F | F | F | T |
| F | F | F | T | F | F | F | F | T |
The statement is contingent since it has both Ts and Fs in its column of the truth table.
In 11 and 12, use truth tables to determine if the following argument forms are valid. You may use a partial truth table if you like. Explain your reasoning.
11.
p <-> q
~q
Therefore: ~p
| p | q | ~p | ~q | p <-> q |
| T | T | F | F | T |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | T | T |
The argument is valid. The premise columns are in bold, and the conclusion is in italic. Since there is no row in which all the premises are true and the conclusion false, the argument is valid. (The only row with all true premises is the last, and that row also has a true conclusion.)
12.
p -> q
~q -> r
Therefore: ~q v ~r
| p | q | r | ~q | ~r | p -> q | ~q -> r | ~q v ~r |
| T | T | T | F | F | T | T | F |
| T | T | F | F | T | T | T | T |
| T | F | T | T | F | F | T | T |
| T | F | F | T | T | F | F | T |
| F | T | T | F | F | T | T | F |
| F | T | F | F | T | T | T | T |
| F | F | T | T | F | T | T | T |
| F | F | F | T | T | T | F | T |
The argument is invalid, as can be seen from the first row in which all the premises (in bold) are true and the conclusion (in italics) is false.
In 13 and 14, use truth tables to determine if the given statement forms are consistent, equivalent, and if either logically entails the other. Explain your answer briefly.
13. ~p -> q , p v ~q
| p | q | ~p | ~q | ~p -> q | p v ~q |
| T | T | F | F | T | T |
| T | F | F | T | T | T |
| F | T | T | F | T | F |
| F | F | T | T | F | T |
Neither statement (first in italic, second in bold) entails the other. They are not equivalent since they differ in truth value on at least one row (the third and fourth). They are consistent because there is at least one row (the first and second) on which they are both true.
14. p , [(q & ~r) -> ~p]
| p | q | r | ~p | ~r | q & ~r | (q & ~r) -> ~p |
| T | T | T | F | F | F | T |
| T | T | F | F | T | T | F |
| T | F | T | F | F | F | T |
| T | F | F | F | T | F | T |
| F | T | T | T | F | F | T |
| F | T | F | T | T | T | T |
| F | F | T | T | F | F | T |
| F | F | F | T | T | F | T |
The first statement (in italics) does not entail the second (in bold) as can be seen from the second row, where the first is true but the second false. And the second does not entail the first as can be seen from the fifth row where the second is true but the first false. They are not equivalent as can be seen from, for example, the second row where they have different truth values. They are consistent, as can be seen from the first row where both are true.
Part Four. Proofs.
Construct proofs for the following arguments. In 15 and 16 you may use all rules of inference and all replacement rules, but not conditional or indirect proof. In 17 and 18 you may also use conditional and indirect proof. In 19 and 20 you MUST use either conditional or indirect proof, or both. You must justify every step.
15.
1. D -> (A v C)
2. D & ~A \ C
3. D [1, simp]
4. A v C [1, 3 MP]
5. ~A [2, simp]
6. C [4, 5 DS]
16.
1. H <-> J
2. ~H \ ~J
3. (H -> J) & (J -> H) [1, BE]
4. J -> H [3, Simp]
5. ~J [2, 4 MT]
17.
1. (A -> B) -> (C -> D)
2. (F -> A) -> (A -> B)
3. A -> (F -> A)
4. A & C \ D v F
5. A [4 simp]
6. F -> A [3, 5 MP]
7. A -> B [2, 6 MP]
8. C -> D [1, 7 MP]
9. C [4 simp]
10. D [8, 9 MP]
11. D v F [10 add]
18.
1. F -> A
2. F & ~A \ D v C
3. F [2 simp]
4. ~A [2 simp]
5. ~F [1, 4 MT]
6. F v (D v C) [3 add]
7. D v C [5, 6 DS]
19.
1. S v P
2. P -> (G & R)
3. ~G
4. P <-> T \ S & ~T
5. |-> ~S [AIP]
6. | P [1, 5 DS]
7. | G & R [2, 6 MP]
8. | G [7 simp]
9. |- G & ~G [3, 8 conj]
10. S [5-9 IP]
11. ~G v ~R [3 add]
12. ~(G & R) [11 dem]
13. ~P [2, 12 MT]
14. (P -> T) & (T -> P) [4 BE]
15. T -> P [14 simp]
16. ~T [13, 15 MT]
17. S & ~T [10, 16 conj]
20.
1. ~M <-> ~N
2. N -> ~M \ ~M
3. |-> M [AIP]
4. | ~N [2, 3 MT, DN]
5. | (~M -> ~N) & ( ~N -> ~M) [1 BE]
6. | ~N -> ~M [5 simp]
7. | ~M [4, 6 MP]
8. |- M & ~M [3, 7 conj]
9. ~M [3-8 IP]