JDEP183H Homework 4a Solution

JDEP183H

Homework Assignment 4a -- Solution

Propositional and Predicate Logic

Assigned: October 26, 2006

Due: 12:30 p.m. November 16, 2006

(Homework 5 minutes late will not be accepted)

1. (30 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) reductio ad absurdum:

(d) (4 points) modus ponens:

(e) (4 points) modus tollens:

(f) (4 points) disjunctive syllogism:

(g) (4 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 reductio ad absurdum: . First we prove that , and then we prove that .

To prove that :

1. Hypothesis

2. 1; implication

3. 2; double negation

4. 3; De Morgan

5. 4; identity

6. 5; implication

So, we have proved that .

To prove that :

1. Hypothesis

2. 1; implication

3. 2; identity

4. 3; De Morgan

5. 4; implication

So, we have proved that .

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

Or we can prove it the following way:

implication

double negation

De Morgan

identity

implication

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

1. Hypothesis

2. 1; implication

3. 2; identity

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

(d) 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.

(e) Prove the modus tollens law:

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.

(f) 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.

(g) 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.

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

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 1; simplification

5. 1; simplification

6. 5, 2; modus tollens

7. 6, 4; conjunction

8. 7, 3; modus ponens

Q.E.D.

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

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 1, 2; disjunctive syllogism

5. 3; implication

6. 4, 5; rule 12.b.

Q.E.D.

1. Hypothesis

2. Hypothesis

3. 1; implication

4. 3; commutative

5. 4; double negation

6. 5; implication

7. 6, 2; hypothetical syllogism or trans. of

8. 7; implication

9. 8; double negation

10. 9; associative

11. 10; implication

Q.E.D.

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

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 1; implication

5. 3; implication

6. 5; De Morgan

7. 6; simplification

8. 6; simplification

9. 7, 1; modus ponens

10. 2, 8; modus tollens

11. 9, 10; conjunction

Q.E.D.

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

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. 2, 3; constructive dilemmas

5. 1, 4; hypothetical syllogism of trans. of

6. 5; implication

7. 6; commutative

8. 7; identity

9. 8; implication

Q.E.D.

3. (20 points) (Based on Ross and Wright 1988). Consider the following hypotheses:

If I take the bus or subway, then I will be late for my appointment. If I take a cab, then I will not be late for my appointment and I will be broke. I will be on time for my appointment.

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) I will take a cab.

(b) I will be broke

(d) If I become broke, then I took a cab.

(e) If I take the bus, then I won’t be broke.

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

b = “I take the bus”

s = “I take the subway”

l = “I will be late for appointment”

c = “I take a cab”

r = “I will be broke”

So the hypotheses are:

“If I take the bus or subway, then I will be late for my appointment.”

“If I take a cab, then I will not be late for my appointment and I will be broke.”

“I will be on time for my appointment.”

(a) Does “I will take a cab” 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, “I will take a cab” does not follow.

(b) Does “I will be broke” 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, “I will be broke” 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, “I will not take the subway” is a valid conclusion.

(d) Does “If I become broke, then I took a cab” 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 I become broke, then I took a cab” does not follow.

(e) Does “If I take the bus, then I won’t be broke” 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 I take the bus, then I won’t be broke” is a valid conclusion.

4. (20 points) Use only the rules given in the tables of Logical Equivalences and Implications in your handout 3. Given the following: If , , , and , then .

(a) (10 points) Give a formal proof for the following without using the contradiction approach: If , , , and , then .

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. Hypothesis

5. 1; De Morgan’s

6. 5, 3; disjunctive syllogism

7. 6; double negation

8. 2, 7; modus ponens

9. 8, 4; disjunctive syllogism

Q.E.D.

(b) (10 points) Give a formal proof for the following with the contradiction approach: If , , , and , then .

1. Hypothesis

2. Hypothesis

3. Hypothesis

4. Hypothesis

5. negation of conclusion

6. 5, 6; conjunction

7. 6; double negation

8. 7; De Morgan’s

9. 8; commutative

10. 2, 9; modus tollens

14. 10, 3; conjunction

15. 1, 14; conjunction

16. contradiction 15; negation

Q.E.D.

5. (20 points) For each of the following English statements, first translate it into symbolic notations using quantifiers and predicates, then negate it (and bring the negation inside the quantifiers), and then translate it back to English statements.

First, let us suppose that x is the universe of discourse for “computer scientist”.

(a) (3 points) “Every computer scientist knows how to write a program.”

The translation is:

The negation is:

The translation of the negation is: “There is at least a computer scientist who does not know how to write a program.”

(b) (3 points) “Not all computer scientists can count.”

The translation is:

The negation is:

The translation of the negation is: “All computer scientist can count.”

The translation is:

The negation is:

The translation of the negation is: “Every computer scientist has not been given any ACM Fellow Awards.”

(d) (4 points) “Every computer scientist graduates from a university.”

Here, we have to define the universe of discourse for all universities. Let us suppose the universe of discourse is z.

The translation is:

The negation is:

The translation of the negation is: “There is at least a computer scientist who does not graduate from a university.”

(e) (6 points) “If a computer scientist is very good, he/she will be given the ACM Fellow Award.”

Let us denote “the ACM Fellow Award” as ACMFA.

The translation is:

The negation is:

We can further reduce the above negation, using the implication and De Morgan laws from our Propositional Logic handouts. So, we have:

implication

De Morgan

The translation of the negation is: “There is at least one computer scientist who is very good and will not be given the ACM Fellow Award.”

6. (10 points) Show that the premises “A car in this garage has an engine problem,” and “Every car in this garage has been sold” imply the conclusion “A car which has been sold has an engine problem.” Let be “x is in this garage,” be “x has an engine problem,” and be “x has been sold.” The premises are and . The conclusion is . Fill in the following blanks to complete the proof: