CSCE235 HW2 Solution

CSCE 235

Homework Assignment 2 - Solution

Assigned: January 21, 2009

Due: 12:30 p.m. January 28, 2009

(Homework 5 minutes late will not be accepted)

1. (25 points) Using only the following logical equivalences: double negation, commutative laws, associative laws, distributive laws, idempotent laws, identity laws, negation laws, De Morgan laws, implication, and equivalence, and the following logical implications: addition, and simplification to logically prove the following, an. You can also use “conjunction”. You must show and explain all steps clearly.

(a) (5 points) exportation law:

(b) (5 points) modus ponens:

(d) (5 points) disjunctive syllogism:

(e) (5 points)

(a) Prove the exportation law: . First we prove that , and then we prove that .

To prove that :

1. Hypothesis

2. 1; implication

3. 2; De Morgan

4. 3; associative

5. 4; implication

6. 5; implication

So, we have proved that .

To prove that :

1. Hypothesis

2. 1; implication

3. 2; implication

4. 3; associative

5. 4; De Morgan

6. 5; implication

So, we have proved that .

Thus, we have shown that . Q.E.D.

Or we can prove it the following way:

implication

De Morgan

associative

implication

Thus, we have shown that . Q.E.D.

(b) Prove the modus ponens law: .

1. Hypothesis

2. 1; implication

3. 2; distributive

4. 3; negation laws

5. 4; commutative

6. 5; identity

7. 6; simplification

Thus, we have shown that . Q.E.D.

1. Hypothesis

2. 1; implication

3. 2; distributive

4. 3; negation laws

5. 4; identity

6. 5; simplification

Thus, we have shown that . Q.E.D.

(d) Prove the disjunctive syllogism law:

1. Hypothesis

2. 1; distributive

3. 2; negation laws

4. 3; identity

5. 4; simplification

Thus, we have shown that . Q.E.D.

(e) Prove

1. Hypothesis

2. 1; addition

3. 2; identity

4. 3; negation

5. 4; commutative

6. 5; distributive

7. 6; implication

Thus, we have shown that . Q.E.D.

2. (50 points) Give a formal proof for each of the following using only the rules given in the tables of Logical Equivalences and Implications in your handout 4.

(a) (10 points) If , , , then .

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 1, 3; Disjunctive Syllogism

5. 4; Addition

6. 5; Modus Ponens

Q.E.D.

(b) (10 points) If and , then .

1. Hypothesis

2. Hypothesis

3. 1; Implication

4. 3; Implication

5. 4; Associative

6. 2; Implication

7. 6; Implication

8. 7; Associative

9. 5, 8; Conjunction

10. 9; Distributive

11. 10; De Morgan’s

12. 11; Implication

13. 12; Commutative

Q.E.D.

1. Hypothesis

_2.Hypothesis

3. Hypothesis

4. Hypothesis

5. 1, 2; Modus Ponens

6. 5, 4; Modus Ponens

7. 3; De Morgan’s

8. 6, 7; Disjunctive Syllogism

Q.E.D.

(d) (10 points) If , , , then .

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 3; Implication

5. 4; De Morgan’s

6. 5; Simplification

7. 5; Simplification

8. 1, 6; Modus Ponens

9. 2, 7; Modus Tollens

10. 7, 9; Conjunction

Q.E.D.

(e) (10 points) If , , , then .

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 1, 2; Transitivity of implication hypothetical syllogism

5. 4, 3; Transitivity of implication hypothetical syllogism

6. 5; Implication

7. 6; Identity

Q.E.D.

5. (20 points) Use formal proof to prove the following. Use only proof by contradiction.

(a) (10 points) If , , , then .

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. Negation of Conclusion

5. 4; De Morgan’s

6. 5; Implication

7. 1, 6; Transitivity of implication, Hypothetical Syllogism

8. 7, 2; Transitivity of implication, Hypothetical Syllogism

9. 3, 8; Conjunction

10. contradiction

Q.E.D.

(b) (10 points) If and , then .

1. Hypothesis

2. Hypothesis

3. Negation of Conclusion

4. 3; Implication

5. 4; De Morgan’s

6. 5; Simplification

7. 5; Simplification

8. 6; Simplification

9. 6; Simplification

10. 9, 1; Modus Ponens

11. 9, 2; Modus Ponens

12. 8, 10; Modus Ponens

13. 8, 11; Modus Ponens

14. 12, 13; Conjunction

15. 14, 7; Conjunction

16. Contradiction

Q.E.D.

4. (20 points) (Based on Ross and Wright 1988). Consider the following hypotheses:

If symptom A or symptom B is observed, then test C is performed. If test D is performed, then test C will not be performed and whether the patient has illness E will be known. Test C is not performed.

Which of the following conclusions must follow, i.e., can be inferred from the above hypotheses? Justify your answers. (Use a logical proof to prove that a conclusion follows; and may use a line of a truth table to prove that a conclusion does not follow.)

(a) Test D is performed.

(b) Whether the patient has illness E will be known

(d) If whether the patient has illness E is known, then test D was performed.

(e) If symptom A is observed, then whether the patient has illness E will not be known.

First, let us represent the problem in the following manner:

a = “Symptom A is observed”

b= “Symptom B is observed”

c = “Test C is not performed” (Note that ‘c’ here is not “contradiction”!)

d = “Test D is performed”

e = “Whether the patient has illness E will be known”

So the hypotheses are:

“If symptom A or symptom B is observed, then test C is performed.”

“If test D is performed, then test C will not be performed and whether the patient has illness E will be known.”

“Test C is performed.”

(a) Does “Test D is performed” follow?

So, is this true: If , , and , then ?

No. It is not true. Basically, we have . Consider the following:


0	0	0	0	0	0	1	0	1	1	1	0	1	1	0	0
Step					1	2	1	4	3	1	2	5	1	6	1

The above scenario shows that is not true when all the propositions are false (column with step ‘6’). Therefore, “Test D is performed” does not follow.

(b) Does “Whether the patient has illness E will be known” follow?

So, is this true: If , , and , then ?

No. It is not true. Basically, we have . Consider the following:


0	0	0	0	0	0	1	0	1	1	1	0	1	1	0	0
Step					1	2	1	4	3	1	2	5	1	6	1

The above scenario shows that is not true when all the propositions are false (column with step ‘6’). Therefore, “Whether the patient has illness E will be known” does not follow.

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 1, 3; modus tollens

5. 4; De Morgan

6. 5; simplification

So, we have proved that “If given , , and , then .” Yes, “Test C is performed” is a valid conclusion.

(d) Does “If whether the patient has illness E is known, then test D was performed” follow? So, given , , and , can we infer ?

No. It is not true. Basically, we have . Consider the following:


0	0	0	0	1	0	1	0	1	1	1	0	1	1	0	0
Step					1	2	1	4	3	1	2	5	1	6	1

The above scenario shows that is not true when proposition is true and all other propositions are false (column with step ‘6’). Therefore, “If whether the patient has illness E is known, then test D was performed” does not follow.

(e) Does “If symptom A is observed, then whether the patient has illness E will not be known” follow? So, given , , and , can we infer ?

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. negation of conclusion

5. 4; implication

6. 5; De Morgan

7. 6; simplification

8. 1, 3; modus tollens

9. 8; De Morgan

10. 9; simplification

11. 10, 7; conjunction

12. contradiction

So, we have proved that “If given , , and , then .” Yes, “If symptom A is observed, then whether the patient has illness E will not be known” is a valid conclusion.