Test Four

Part I: Use the letters provided to symbolize the following sentences. (26 points)

Let: Lx = x is liberal. Dx = x is a Democrat. a = Arbuckle.

Cx = x is conservative. Rx = x is a Republican. b = Burke.

Mx = x is moderate. Gx = x supports the Greens. c = Clay.

1. Arbuckle is a moderate Democrat while Clay is a conservative Republican.

 

 

2. If Burke is neither liberal nor conservative, then there are moderate Republicans.

 

 

3. Liberals are not conservatives.

 

 

4. Some conservatives are not Republicans.

 

 

5. Not all liberals support the Greens.

 

 

6. Any Democrat who is not conservative is either moderate or liberal.

 

 

7. If no Republicans support the Greens, then Burke is not a Republican.

 

 

8. Only liberals and moderates support the Greens.

Part II: More symbolization. Use the translation schemas above, and these:

Wx = x won the election.

Vxy = x voted y. (15 points)

1. If Arbuckle voted for Clay, then Burke did not win the election.

 

 

2. If Burke voted for anyone, then she voted for herself.

 

 

3. Arbuckle won the election only if all liberal Democrats voted for him.

 

 

4. Every Republican who voted for Clay voted for Burke, too.

 

 

5. No liberals voted for any conservatives.

 

 

Part III: Syntax. (10 points)

[ (x) R x y · F a ] … ($y)( B y v R y x )

In the above symbolic sentence:

(i) Circle any free variables.

(ii) Underline the scope of any universal quantifier.

(iii) Double underline the scope of any existential quantifier.

(iv) List all the names, if any:

(v) What kind of sentence is it (What is its logical form)?

 

Part IV: Identify any errors in this "proof" in the space at the right, then treat the uncorrected line as correct for the rest of the proof. (25 points)

1. (y)Ay … ($x)~Dx

2. (x)(~Ax … Bx)

3. (x)~Bx v (x)(Cx … Dx)

4. (x)(Dx v Fx) … ($x)~Dx /\ ($x)(Ax • Bx)

5. Æ ~($x)Ax AP for IP

6. ~Ax EI 5

7. ~Ax … Bx UI 2

8. Bx MP 6, 7

9. (x)Bx UG 9

10. (x)(Cx … Dx) DS 3, 9

11. ~Ac EI 5

12. ($y)~Ay EG 11

13. ~(y)Ay CQ 12

14. ($x)~Dx … (y)Ay Comm 1

15. ~($x)~Dx MT 14, 13

16. ($x)Dx DN 15

17. Db UI 16

18. Db v Fb Add 17

19. (Db v Fb) … ($x)~Dx UI 4

20. ($x)~Dx MP 18, 19

21. ($x)Dx • ($x)~Dx Conj 16, 20

22. ($x)Ax IP 4-22

23. ($x)Ax • ($x)Dx Conj 22, 16

24. ($x)(Ax • Dx) Q. Dist 23

 

Part V: Using any of the available rules of inference and replacement, including CP and IP, prove these arguments are valid. (12 points each)

A. 1. (x)(Ax … Fx) B. 1. ($x)(Mx • Bx)

2. ($x)Fx … (x)~Hx | 2. ($x)(Ax v Lx) … ~(x)Ox

3. (x)(Hx v Kx) | 3. ($x)Cx … (x)(Mx … Tx)

4. ($x)(Ax • Gx) /\ (x)Kx | 4. (x)(Ox v Cx) /\ ($x)Ax … ($x)Tx

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Test Four: Answers

Part I:

1. (Ma • Da) • (Cc • Rc) 5. ~(x)(Lx … Gx) [or: ($x)(Lx • ~Gx)]

2. (~Lb • ~Cb) … ($x)(Mx • Rx) 6. (x)[(Dx • ~Cx) … (Mx v Lx)]

3. (x)(Lx … ~Cx) [or: ~($x)(Lx • Cx)] 7. ~($x)(Rx • Gx) … ~Rb

4. ($x)(Cx • ~Rx) 8. (x)[Gx … (Lx v Mx)]

Part II:

    1. Vac … ~Wb 4. (x)[(Rx • Vxc) … Vxb]
    2. ($x)Vbx … Vbb 5. ~($x)[Lx • ($y)(Cy • Vxy)]

3. Wa … (x)[(Lx • Dx) … Vxa] [or: (x)(Lx … (y)(Cy … ~Vxy)) ]

Part III:

[ (x) R x y · F a ] … ($y)( B y v R y x )

The first "y" and the last "x" are the free variables;

"a" is the only name;

"…" is the main connective.

Part IV: Identify any errors

1. (y)Ay … ($x)~Dx

2. (x)(~Ax … Bx)

3. (x)~Bx v (x)(Cx … Dx)

4. (x)(Dx v Fx) … ($x)~Dx /\ ($x)(Ax • Bx)

5. Æ ~($x)Ax AP for IP

6. ~Ax EI 5 X — The negation is in the way; Must CQ first

7. ~Ax … Bx UI 2 ˆ - OK

8. Bx MP 6, 7 ˆ - OK

9. (x)Bx UG 8 X — "x" was free on a line that came by EI (line 6)

10. (x)(Cx … Dx) DS 3, 9 X - ~(x)~Bx is needed for the DS

11. ~Ac EI 5 X - Negation is still in the way.

12. ($y)~Ay EG 11 ˆ - OK

13. ~(y)Ay CQ 12 ˆ - OK

14. ($x)~Dx … (y)Ay Comm 1 X — Cannot commute conditionals.

15. ~($x)~Dx MT 14, 13 ˆ - OK

16. ($x)Dx DN 15 X — The ~’s apply to different parts; use CQ

17. Db UI 16 X — The inference is OK but the rule is EI

18. Db v Fb Add 17 ˆ - OK

19. (Db v Fb) … ($x)~Dx UI 4 X — Cannot UI part of a line; Line 4 is a conditional.

20. ($x)~Dx MP 18, 19 ˆ - OK

21. ($x)Dx • ($x)~Dx Conj 16, 20 ˆ - (The inference is OK, but…)

22. ($x)Ax IP 5-21 X - … Line 21 is not a contradiction

23. ($x)Ax • ($x)Dx Conj 22, 16 X - Cannot use line 16 out of the subproof.

24. ($x)(Ax • Dx) Q. Dist 23 X — Nice try, but there’s no such rule.

Part V: Proofs

A. 1. (x)(Ax … Fx) B. 1. ($x)(Mx • Bx)

2. ($x)Fx … (x)~Hx | 2. ($x)(Ax v Lx) … ~(x)Ox

3. (x)(Hx v Kx) | 3. ($x)Cx … (x)(Mx … Tx)

4. ($x)(Ax • Gx) /\ (x)Kx | 4. (x)(Ox v Cx) /\ ($x)Ax … ($x)Tx

5. Ab • Gb EI 4 (use a name) 5. Æ ($x)Ax AP for CP

6. Ab Simp 5 6. Aa EI 5 (a name)

7. Ab … Fb UI 1 7. Aa v La Add 6

8. Fb MP 6, 7 8. ($x)(Ax v Lx) EG 7

9. ($x)Fx EG 8 9. ~(x)Ox MP 2, 8

10. (x)~Hx MP 2, 9 10. ($x)~Ox CQ 9

11. ~Hx UI 10 (use a variable!) 11. ~Ob EI 10 (a new name)

12. Hx v Kx UI 3 12. Ob v Cb UI 4

13. Kx DS 11, 12 13. Cb DS 11, 12

14. (x)Kx UG 13 14. ($x)Cx EG 13

-------- An indirect proof -------- 15. (x)(Mx … Tx) MP 14, 3

5. Æ ~(x)Kx AP for IP 16. Mc • Bc EI 1 (another name)

6. ($x)~Kx CQ 5 17. Mc Simp 16

7. ~Ke EI 6 18. Tc MP 16, 17

8. He v Ke UI 3 19. ($x)Tx EG 18

9. He DS 7, 8 20. ($x)Ax … ($x)Tx CP 5 -19

10. ($x)Hx EG 9

11 ($x)Fx … ~($x)Hx CQ 2

12. ~($x)Fx MT, DN 12, 11

13. Ag • Gg EI 4 (Use a new name)

14. Ag Simp 13

15. Ag … Fg UI 1

16. Bg MP 14, 15

17. ($x)Fx EG 16

18 ($x)Fx • ~($x)Fx Conj 16, 17

19. (x)Kx IP 5 -18.