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Math 310Exam 1 - Practice Problem Solutions

1. Translate the following English statements into symbolic form (use propositional sentences).

(a) I will not pass this class unless I go to class every day anddo all of the homework exercises.

First we define variables.p: I pass this class.g: I go to class everyday.h: I do all the homework exercises.

Then the statement translates as: (¬g? ¬h)→ ¬por as:p→(g?h) (b) I lock the doors and close the windows whenever I leave to go to work. First we define variables.d: I lock the doors.c: I close the windows.w: I leave to go to work.

Then the statement translates as:w→(d?c)

(c) Getting up on time and getting ready quickly is sufficient for arriving at work on time. First we define variables.g: I get up on time.q: I get ready quickly.w: I arrive at work on time.

Then the statement translates as: (g?q)→w

(d) Practicing an hour a day and getting private lessons twice a week is necessary for playing in the wind ensemble.

First we define variables.p: I practice an hour a day.l: I get private lessons twice a week.w: I play in the wind

ensemble.

Then the statement translates as:w→(p?l)

2. Use truth tables to determine whether or not the followingpairs of statements are logically equivalent.

(a) [(p?q)→r] and (p→r)?(q→r) pqrp?q(p?q)→r TTTTT TTFTF TFTFT TFFFT FTTFT FTFFT FFTFT FFFFT pqrp→rq→r(p→r)?(q→r)

TTTTTT

TTFFFF

TFTTTT

TFFFTF

FTTTTT

FTFTFF

FFTTTT

FFFTTT

Since the last columns of there truth tables are not identical, these two propositions are not logically equivalent.

(b)p?(q?r) and (p?q)?(p?r) pqrq?rp?(q?r) TTTTT TTFTT TFTTT TFFFF FTTTF FTFTF FFTTF FFFFF pqrp?qp?r(p?r)?(p?r)

TTTTTT

TTFTFT

TFTFTT

TFFFFF

FTTFFF

FTFFFF

FFTFFF

FFFFFF

Since the last columns of there truth tables are identical, these two propositions are logically equivalent.

(c)p?(q? ¬r) and (p?q)? ¬(¬p?r) pqr¬rq? ¬rp?(q? ¬r)

TTTFTT

TTFTTT

TFTFFF

TFFTTT

FTTFTF

FTFTTF

FFTFFF

FFFTTF

pqrp?q¬p¬p?r¬(¬p?r)(p?q)? ¬(¬p?r)

TTTTFTFT

TTFTFFTT

TFTFFTFF

TFFFFFTT

FTTFTTFF

FTFFTTFF

FFTFTTFF

FFFFTTFF

Since the last columns of there truth tables are identical, these two propositions are logically equivalent.

3. Use truth tables to determine which of the following statements are tautologies:

(a) (p?q)↔[(p?q)? ¬(p?q)] pqp?qp?qp?q¬(p?q)(p?q)? ¬(p?q)(p?q)↔(p?q)? ¬(p?q)

TTFTTFFT

TFTTFTTT

FTTTFTTT

FFFFFTFT

Since every entry in the final column isT, this proposition is a tautology. (b) (p→q)↔[¬(p? ¬q)] pqp→q¬qp? ¬q¬(p? ¬q)(p→q)↔[¬(p? ¬q)]

TTTFFTT

TFFTTFT

FTTFFTT

FFTTFTT

Since every entry in the final column isT, this proposition is a tautology. (c)p→(¬q?r? ¬r) pqr¬q¬r¬q? ¬r¬q?r? ¬rp→(¬q?r? ¬r)

TTTFFFTT

TTFFTTTT

TFTTFTTT

TFFTTTTT

FTTFFFTT

FTFFTTTT

FFTTFTTT

FFFTTTTT

Since every entry in the final column isT, this proposition is a tautology.

4. Use propositional rules of inference to prove the the following statements are tautologies:

(a) [¬p?(p?q)]→q

We proceed by constructing a 2-column proof:

Statement

Reason

[¬p?(p?q)]→qGiven [(¬p?p)?(¬p?q)]→qDistributive Law [F?(¬p?q)]→qNegation Law (¬p?q)→qIdentity Law (¬p→q)?(q→q)Established Conditional Equivalence (¬p→q)?(¬q?q)Conditional to Disjunction (¬p→q)?TNegation Law

TDomination Law

The proof given in the table above verifies that the original proposition is always true. (b)¬p→(p→q)

We proceed by constructing a 2-column proof:

Statement

Reason

¬p→(p→q)Given

p?(p→q)Conditional to Disjunction p?(¬p?q)Conditional to Disjunction (p? ¬p)?qAssociative Law

T?qNegation Law

TDomination Law

The proof given in the table above verifies that the original proposition is always true.

5. Use propositional rules of inference to show that each pair of logical statements are logically equivalent.

(a) (r?p)→(r?q) andr?(p→q).

Statement

Reason

(r?p)→(r?q)Given

¬(r?p)?(r?q)Conditional to Disjunction

(¬r? ¬p)?(r?q)DeMorgan"s Law [(¬r? ¬p)?r]?qAssociative Law [r?(¬r? ¬p)]?qCommutative Law [(r? ¬r)?(r? ¬p)]?qDistributive Law [T?(r? ¬p)]?qNegation Law [(r? ¬p)]?qIdentity Law r?(¬p?q)Associative Law r?(p→q)Conditional to Disjunction

Thus (r?p)→(r?q)≡r?(p→q)

(b)p→qand [(p? ¬q)→ ¬p]

Statement

Reason

[(p? ¬q)→ ¬p]Given

¬(p? ¬q)? ¬pConditional to Disjunction

(¬p?q)? ¬pDeMorgan"s Law (q? ¬p)? ¬pCommutative Law q?(¬p? ¬p)Associative Law q? ¬pIdempotent Law

¬p?qCommutative Law

p→qConditional to Disjunction

Thusp→q≡[(p? ¬q)→ ¬p]

6. Determine whether or not the following statements are satisfiable.

(a) (p?q? ¬r)?(p? ¬q?r)?(¬p? ¬q?r)

Consider the truth assignmentp:T,q:F,andr:T.

Thenp?q? ¬rhas truth valueT?F?F, so this part of the proposition is true. Similarly,p? ¬q?rhas truth valueT?T?T, so this part of the proposition is true. Similarly,¬p? ¬q?rhas truth valueF?T?T, so this part of the proposition is true.

Thus the the entire statement (the conjunction of the previous statements) is true with the given truth value

assignment. (b) (p?q? ¬r?s)?(p? ¬q?r? ¬s)?(¬p? ¬q?r? ¬s)

Consider the truth assignmentp:T,q:F,r:Tands:F.

Thenp?q? ¬r?shas truth valueT?F?F?F, so this part of the proposition is true. Similarly,p? ¬q?r? ¬shas truth valueT?T?T?T, so this part of the proposition is true. Similarly,¬p? ¬q?r? ¬shas truth valueF?T?T?T, so this part of the proposition is true.

Thus the the entire statement (the conjunction of the previous statements) is true with the given truth value

assignment.

7. Suppose that there is a certain town in which all the men either shave themselves, or they are shaved by the town

barber Figaro (who is male). Suppose Vinny is too cheap to have the barber shave him. LetS(x,y) be the two variable

predicate "personxis shaved by person "y". First translate the given statement into English, then determine the truth

value of the statement and justify your answer. (a)?x(S(x,Figaro)?S(x,x)) Note that the implied domain for bothxandyis the men who live in this town. Translation: Every man in the town is either shaved by the barber Figaro or shaves himself.

Value: True (this is the given statement above).

(b)?x(¬S(x,x)→S(x,Figaro)) Translation: Every man in the town that does not shave himself is shaved by the barber Figaro. Value: True (since there are only two possibilities, one of them must hold). (c)?x(S(x,Figaro)→ ¬S(x,x)) Translation: Every man in the town that is shaved by the barber Figaro does not shave himself. Value: False (the counterexample is Figaro, who shaves himself and is also shaved by Figaro). (d)?y?x(S(x,y)) Translation: There is a man who shaves every man in the town. Value: False (Since Vinny shaves himself, there is noone whoshaves everyone). (e)?!x(S(x,x)?S(x,Figaro)) Translation: There is exactly one man who both shaves himself and is shaved by the barber Figaro. Value: True (the example is Figaro, who shaves himself and isalso shaved by Figaro).

8. Translate each of the following statements into quantified predicate form. Make sure to define each predicate used and

state the domain of each variable. (a) At least one person in this neighborhood watches television on Monday but not on Wednesday. Letxbe the set of all people. Letybe the set of days of the week. LetN(x) denote: personxlives in this neighborhood. LetW(x,y) denote: personxwatches television on dayy. Translation:?x(N(x)?W(x,Monday)? ¬W(x,Wednesday)). (b) There is a person who has run a marathon in every state in the United States of America. Letxbe the set of all people. Letybe the set of states in the U.S. LetR(x,y) denote: personxhas run a marathon in statey.

Translation:?x?y(R(x,y)).

9. Write the negation of each of the following statements (First write each statement symbolically, then negate the symbolic

statement, and finally, translate the negation back into plain English). (a) Everyone who took their driver"s exam today passed the exam.

Letxbe the set of all people,T(x) the predicate: "Personxtook their driver"s exam.", andP(x) the predicate:

"Personxpassed their driver"s exam." Then the translation of this statement in symbolic form is:?x(T(x)→P(x)). Negating this statement, we get¬?x(T(x)→P(x))≡ ?x¬(T(x)→P(x)) ≡ ?x¬(¬T(x)?P(x))≡ ?x(T(x)? ¬P(x)).

Translating back to English, this is: "There is some person who took their driver"s exam but did not pass their

driver"s exam." (b) Some people like bowling and tennis.

Letxbe the set of all people,B(x) the predicate: "Personxlikes bowling.", andT(x) the predicate: "Personx

likes tennis." Then the translation of this statement in symbolic form is:?x(B(x)?T(x)).

Negating this statement, we get¬?x(B(x)?T(x))≡ ?x¬(B(x)?T(x))≡ ?x(¬B(x)? ¬T(x)).

Translating back to English, this is: "Everyone either dislikes Bowling or dislikes tennis." (c) If everyone passed the exam then everyone studied for theexam.

Letxbe the set of all people,P(x) the predicate: "Personxpassed the exam.", andS(x) the predicate: "Person

xstudied for the exam." Then the translation of this statement in symbolic form is: (?xP(x))→(?xS(x)).

Negating this statement, we get¬[(?xP(x))→(?xS(x))]≡ ¬[(¬?xP(x))?(?xS(x))]≡ ?xP(x)? ¬(?xS(x)).

≡ ?xP(x)? ?x¬S(x)

Translating back to English, this is: "Everyone passed the exam but some person did not study for the exam."

10. Negate each of the statements (your answer should be in symbolic form)

(a)?x?y[((x >0)?(y <0))→(xy≥0)] ¬?x?y[((x >0)?(y <0))→(xy≥0)]≡ ?x?y¬[((x >0)?(y <0))→(xy≥0)] ≡ ?x?y¬[¬((x >0)?(y <0))?(xy≥0)]≡ ?x?y[((x >0)?(y <0))? ¬(xy≥0)] ≡ ?x?y[((x >0)?(y <0))?(xy <0)] (b)?x?y?z[(F(x,y)?G(x,z))→H(y,z)] ¬?x?y?z[(F(x,y)?G(x,z))→H(y,z)]≡ ?x?y?z¬[(F(x,y)?G(x,z))→H(y,z)] ≡ ?x?y?z¬[¬(F(x,y)?G(x,z))?H(y,z)] ≡ ?x?y?z[(F(x,y)?G(x,z))? ¬H(y,z)] (c)?x?!y[S(x,y)? ¬R(x,y)] ¬?x?!y[S(x,y)? ¬R(x,y)]≡ ?x¬?!y[S(x,y)? ¬R(x,y)]

≡ ?x[?y¬[S(x,y)? ¬R(x,y)]? ?y1[S(x,y1)? ¬R(x,y1)]? ?y2[(y1?=y2)?[S(x,y2)? ¬R(x,y2)]]]

≡ ?x[?y[¬S(x,y)?R(x,y)]]? ?y1[S(x,y1)? ¬R(x,y1)]? ?y2[(y1?=y2)?[S(x,y2)? ¬R(x,y2)]]

11. Letx,yandzbe integers. Determine the truth value of each of the following.

(a)?x?y(x+y= 1) TRUE

To see this, note that, given an integerx, if we lety= 1-x, thenyis also an integer, andx+y=x+(1-x) = 1.

(b)?x?y?z(xy < z) TRUE To see this, note that, given arbitrary integersxandy, if we letz=xy+ 1, thenxy < z. (c)?z?x?y(xy < z) FALSE To see this, note that unlike in part (b), given the order of quantification, we must choosezfirst.

Givenz, if we letx=z+ 1 andy= 1 thenxy=z+ 1> z.

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