Fall 2008
Philosophy 3200
1) Prove the following Quantifier Negation rule. You may use
any primitive or derived rules, except, of course, for any of
the Quantifier Negation rules.
(4 points)
a)
(∀x)~Fx
~(∃x)Fx
1.
2.
3.
4.
5.
6.
7.
| (∀x)~Fx
| | (∃x)Fx
| | a|Fa
| | |~Fa
| | |⊥
| |⊥
|~(∃x)Fx
P
A
A
1 ∀E
3,4 ⊥I
2, 3-5 ∃E
Note: Where I use falsum (‘⊥’), Teller uses ‘A & ~A’. You may
use whatever you want to indicate a contradiction, as long as
you don’t violate the isolation of the relevant individual used
in ∃E.
Exam #6
2) Provide derivations that establish the validity of the
following arguments. You may use any of the primitive or
derived rules. (9 points each)
a)
(∀x)(Ex ⊃ Lxj)
(∀y)~Lyj
(∀z)(Ez v Kfz)
(∀x)Kfx
1.
2.
3.
4.
5.
6.
7.
8.
9.
|(∀x)(Ex ⊃ Lxj)
|(∀y)~Lyj
|(∀z)(Ez v Kfz)
|Eâ ⊃ Lâj
|~Lâj
|Eâ v Kfâ
|~Eâ
|Kfâ
|(∀x)Kfx
P
P
P
1 ∀E
2 ∀E
3 ∀E
4,5 DC
6,7 vE
8 ∀I
Fall 2008
Philosophy 3200
b) (∀z)[Kzz ⊃ (Mz & Nz)]
(∃z)~Nz
(∃x)~Kxx
1. |(∀z)[Kzz ⊃ (Mz & Nz)]
2. |(∃z)~Nz
3. | a |~Na
4. | |Kaa ⊃ (Ma & Na)
5. | | |Kaa
6. | | |Ma & Na
7. | | |Na
8. | | |~Na
9. | |~Kaa
10. | |(∃x)~Kxx
11. |(∃x)~Kxx
c)
d) (∃x)Bx ⊃ (∀y)Ay
(∀w)(Bw ⊃ Aw)
P
P
A
1 ∀E
A
4,5 ⊃E
6 &E
3R
5-8 ~I
9 ∃I
2, 3-10 ∃E
(∀z)(~Lz ⊃ ~Bz)
(∃x)~Lx ⊃ (∃y)~By
1.
2.
3.
4.
5.
6.
7.
8.
|(∀z)(~Lz ⊃ ~Bz)
| |(∃x)~Lx
| | a |~La
| | |~La ⊃ ~Ba
| | |~Ba
| | |(∃y)~By
| |(∃y)~By
|(∃x)~Lx ⊃ (∃y)~By
Exam #6
1. |(∃x)Bx ⊃ (∀y)Ay
2. | |~(∀w)(Bw ⊃ Aw)
3. | |(∃w)~(Bw ⊃ Aw)
4. | | a |~(Ba ⊃ Aa)
5. | | |Ba & ~Aa
6. | | |Ba
7. | | |(∃x)Bx
8. | | |(∀y)Ay
9. | | |Aa
10. | | |~Aa
11. | | |⊥
12. | |⊥
13. |(∀w)(Bw ⊃ Aw)
or
P
A
A
1 ∀E
3,4 ⊃E
5 ∃I
2, 3-6 ∃E
2-7 ⊃I
1.
2.
3.
4.
5.
6.
7.
|(∃x)Bx ⊃ (∀y)Ay
| |Ba
| |(∃x)Bx
| |(∀y)Ay
| |Aa
|Bâ ⊃ Aâ
|(∀w)(Bw ⊃ Aw)
P
A
2 ∃I
1,3 ⊃E
4 ∀∃
2-5 ⊃I
6 ∀I
P
A
2 QN
A
4C
5 &E
6 ∃I
1,7 ⊃E
8 ∀E
5 &E
9,10 ⊥I
3, 4-11 ∃E
2-12 RD
Fall 2008
Philosophy 3200
3) Provide derivations that establish the validity of the
following arguments. You may use any of the primitive or
derived rules. (12 points each)
a)
(∃x)Fx ⊃ (∀x)(Gx ⊃ Fx)
(∃x)Hx ⊃ (∀x)(Fx ⊃ Hx)
(∃x)(Fx & Hx)
(∀x)(Gx ⊃ Hx)
1. |(∃x)Fx ⊃ (∀x)(Gx ⊃ Fx)
2. |(∃x)Hx ⊃ (∀x)(Fx ⊃ Hx)
3. |(∃x)(Fx & Hx)
4. | a |Fa & Ha
5. | | |Ge
6. | | |Fa
7. | | |(∃x)Fx
8. | | |(∀x)(Gx ⊃ Fx)
9. | | |Ge ⊃ Fe
10. | | |Ha
11. | | |(∃x)Hx
12. | | |(∀x)(Fx ⊃ Hx)
13. | | |Fe ⊃ He
14. | | |Fe
15. | | |He
16. | |Gê ⊃ Hê
17. | |(∀x)(Gx ⊃ Hx)
18. |(∀x)(Gx ⊃ Hx)
P
P
P
A
A
4 &E
6 ∃I
1,7 ⊃E
8 ∀E
4 &E
10 ∃I
2,11 ⊃E
12 ∀E
5,9 ⊃E
13,14 ⊃E
5-15 ⊃I
16 ∀I
3, 4-17 ∃E
Exam #6
Fall 2008
b)
Philosophy 3200
(∀x)(∀y)(Gx ⊃ Hy)
(∃x)[Fx & (∃y)(Gy & ~Hx)]
(∃x)[Fx & (∀z)~Gz]
1. |(∀x)(∀y)(Gx ⊃ Hy)
2. |(∃x)[Fx & (∃y)(Gy & ~Hx)]
3. | a |Fa & (∃y)(Gy & ~Ha)
4. | |(∃y)(Gy & ~Ha)
5. | | b |Gb & ~Ha
6. | | |(∀y)(Gb ⊃ Hy)
7. | | |Gb ⊃ Ha
8. | | |Gb
9. | | |Ha
10. | | |~Ha
11. | | |(∃x)[Fx & (∀z)~Gz]
12. | |(∃x)[Fx & (∀z)~Gz]
13. |(∃x)[Fx & (∀z)~Gz]
Exam #6
or:
P
P
A
3 &E
A
1 ∀E
6 ∀E
5 &E
7,8 ⊃E
5 &E
9,10 CD
4, 5-11 ∃E
2, 3-12 ∃E
1. |(∀x)(∀y)(Gx ⊃ Hy)
2. |(∃x)[Fx & (∃y)(Gy & ~Hx)]
3. | a |Fa & (∃y)(Gy & ~Ha)
4. | |(∃y)(Gy & ~Ha)
5. | | b |Gb & ~Ha
6. | | | |Ge
7. | | | |(∀y)(Ge ⊃ Hy)
8. | | | |Ge ⊃ Ha
9. | | | |Ha
10. | | | |~Ha
11. | | |~Gê
12. | | |(∀z)~Gz
13. | | |Fa
14. | | |Fa & (∀z)~Gz
15. | | |(∃x)[Fx & (∀z)~Gz]
16. | |(∃x)[Fx & (∀z)~Gz]
17. |(∃x)[Fx & (∀z)~Gz]
P
P
A
3 &E
A
A
1 ∀E
7 ∀E
6,8 ⊃E
5 &E
6-10 ~I
11 ∀I
3 &E
12,13 &I
14 ∃I
4, 5-15 ∃E
2, 3-16 ∃E
Fall 2008
c)
Philosophy 3200
~(∀x)[Fx ⊃ (Gx & (∃y)Hy)]
~(∃y)(∀x)[Fx ⊃ (Gx & Hy)]
1. |~(∀x)[Fx ⊃ (Gx & (∃y)Hy)]
2. | |(∃y)(∀x)[Fx ⊃ (Gx & Hy)]
3. | | a |(∀x)[Fx ⊃ (Gx & Ha)]
4. | | |(∃x)~[Fx ⊃ (Gx & (∃y)Hy)]
5. | | | b |~[Fb ⊃ (Gb & (∃y)Hy]
6. | | | |Fb & ~(Gb & (∃y)Hy)
7. | | | |Fb
8. | | | |~(Gb & (∃y)Hy)
9. | | | |~Gb v ~(∃y)Hy
10. | | | |Fb ⊃ (Gb & Ha)
11. | | | |Gb & Ha
12. | | | |Gb
13. | | | |~(∃y)Hy
14. | | | |Ha
15. | | | |(∃y)Hy
16. | | | |⊥
17. | | |⊥
18. | |⊥
19. |~(∃y)(∀x)[Fx ⊃ (Gx & Hy)]
P
A
A
1 QN
A
5C
6 &E
6 &E
8 DM
3 ∀E
7,10 ⊃E
11 &E
9,12 vE
11 &E
14 ∃I
13,15 ⊥I
4, 5-16 ∃E
2, 3-17 ∃E
2-18 ~I
Exam #6
Or you could try and do it by working backwards
(there are a lot of variations on this strategy, e.g., you
could work backwards one more step from line 13 to
‘Fa & ~(Ga & Hê)’ and assume ‘Ga & He’ at line 6):
1. |~(∀x)[Fx ⊃ (Gx & (∃y)Hy)]
2. |(∃x)~[Fx ⊃ (Gx & (∃y)Hy)]
3. | a |~[Fa ⊃ (Ga & (∃y)Hy)]
4. | |Fa & ~(Ga & (∃y)Hy)
5. | |Fa
6. | | |Fa ⊃ (Ga & He)
7. | | |Ga & He
8. | | |Ga
9. | | |He
10. | | |(∃y)Hy
11. | | |Ga & (∃y)Hy
12. | | |~(Ga & (∃y)Hy)
13. | |~[Fa ⊃ (Ga & Hê)]
14. | |(∃x)~[Fx ⊃ (Gx & Hê)]
15. | |~(∀x)[Fx ⊃ (Gx & Hê)]
16. | |(∀y)~(∀x)[Fx ⊃ (Gx & Hy)]
17. | |~(∃y)(∀x)[Fx ⊃ (Gx & Hy)]
18. |~(∃y)(∀x)[Fx ⊃ (Gx & Hy)]
P
1 QN
A
3C
4 &E
A
5,6 ⊃E
7 &E
7 &E
9 ∃I
8,10 &I
4 &E
6-12 ~I
13 ∃I
14 QN
15 ∀I
16 QN
2, 3-17 ∃E
Fall 2008
d)
Philosophy 3200
(∀x)(∀y)(∀z)(Dx ≡ ~Hxzy)
(∃x)(∃y)(∀z)[Dx & (Fyz ⊃ Hxyz)
~(∀w)(∀x)(∃y)(Fxy v Hwxy)
1. |(∀x)(∀y)(∀z)(Dx ≡ ~Hxzy)
2. |(∃x)(∃y)(∀z)[Dx & (Fyz ⊃ Hxyz)
3. | a |(∃y)(∀z)[Da & (Fyz ⊃ Hayz)
4. | | b |(∀z)[Da & (Fbz ⊃ Habz)]
5. | | |Da & (Fbê ⊃ Habê)
6. | | |(∀y)(∀z)(Da ≡ ~Hazy)
7. | | |(∀z)(Da ≡ ~Hazê)
8. | | |Da ≡ ~Habê
9. | | |Da
10. | | |~Habê
11. | | |Fbê ⊃ Habê
12. | | |~Fbê
13. | | |~Fbê & ~Habê
14. | | |~(Fbê v Habê)
15. | | |(∀y)~(Fby v Haby)
16. | | |~(∃y)(Fby v Haby)
17. | | |(∃x)~(∃y)(Fxy v Haxy)
18. | | |~(∀x)(∃y)(Fxy v Haxy)
19. | | |(∃w)~(∀x)(∃y)(Fxy v Hwxy)
20. | | |~(∀w)(∀x)(∃y)(Fxy v Hwxy)
21. | |~(∀w)(∀x)(∃y)(Fxy v Hwxy)
22. |~(∀w)(∀x)(∃y)(Fxy v Hwxy)
Exam #6
or:
P
P
A
A
4 ∀E
1∀E
6 ∀E
7 ∀E
5 &E
8,9 ≡E
5,9 ⊃E
10,11 DC
10,12 &I
13 DM
14 ∀I
15 QN
16 ∃I
17 QN
18 ∃I
19 QN
3, 4-20 ∃E
2, 3-21 ∃E
1. |(∀x)(∀y)(∀z)(Dx ≡ ~Hxzy)
2. |(∃x)(∃y)(∀z)[Dx & (Fyz ⊃ Hxyz)
3. | a |(∃y)(∀z)[Da & (Fyz ⊃ Hayz)
4. | | b |(∀z)[Da & (Fbz ⊃ Habz)]
5. | | | |(∀w)(∀x)(∃y)(Fxy v Hwxy)
6. | | | |(∀x)(∃y)(Fxy v Haxy)
7. | | | |(∃y)(Fby v Haby)
8. | | | | c |Fbc v Habc
9. | | | | |Da & (Fbc ⊃ Habc)
10. | | | | |(∀y)(∀z)(Da ≡ ~Hazy)
11. | | | | |(∀z)(Da ≡ ~Hazc)
12. | | | | |Da ≡ ~Habc
13. | | | | |Da
14. | | | | |~Habc
15. | | | | |Fbc ⊃ Habc
16. | | | | |~Fbc
17. | | | | |Fbc
18. | | | | |⊥
19. | | | |⊥
20. | | |~(∀w)(∀x)(∃y)(Fxy v Hwxy)
21. | |~(∀w)(∀x)(∃y)(Fxy v Hwxy)
22. |~(∀w)(∀x)(∃y)(Fxy v Hwxy)
P
P
A
A
A
5 ∀E
6 ∀E
A
4 ∀E
1 ∀E
10 ∀E
11 ∀E
9 &E
12,13 ≡E
9 &E
14,15 DC
8,14 vE
16,17 ⊥I
7, 8-18 ∃E
5-19 ~I
3, 4-20 ∃E
2, 3-21 ∃E
Fall 2008
e)
Philosophy 3200
(∃x)[Hx & (∃y)(Dy & Lxy)]
(∃x)[Hx & (∀y)(Dy ⊃ ~Lxy)]
(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]}
1. |(∃x)[Hx & (∃y)(Dy & Lxy)]
P
2. |(∃x)[Hx & (∀y)(Dy ⊃ ~Lxy)]
P
3. | a |Ha & (∃y)(Dy & Lay)
A
4. | |(∃y)(Dy & Lay)
3 &E
5. | | b |Db & Lab
A
6. | | | c |Hc & (∀y)(Dy ⊃ ~Lcy)
A
7. | | | | |~(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]} A
8. | | | | |(∃x)~{Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]} 7 QN
9. | | | | | d |~{Dd ⊃ (∃y)[Hy & (Lyd v ~Lyd)]} A
10. | | | | | |Dd & ~(∃y)[Hy & (Lyd v ~Lyd)]
9C
11. | | | | | |~(∃y)[Hy & (Lyd v ~Lyd)] 10 &E
12. | | | | | |(∀y)~[Hy & (Lyd v ~Lyd)] 11 QN
13. | | | | | |~[Ha & (Lad v ~Lad)]
12 ∀E
14. | | | | | |~Ha v ~(Lad v ~Lad)
13 DM
15. | | | | | |Ha
3 &E
16. | | | | | |~(Lad v ~Lad)
14,15 vE
17. | | | | | |⊥
16 ⊥I
18. | | | | |⊥
8, 9-17 ∃E
19. | | | |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]} 7-18 RD
20. | | |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]} 2, 6-19 ∃E
21. | |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]}
4, 5-20 ∃E
22. |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]}
1, 3-21 ∃E
This could have been shorter by eliminating lines 4-6, do you
see why? Would you have seen it if you used RD as your main
strategy?
Exam #6
Or work backwards from your target sentence once you’ve
dispatched all of the ∃’s, giving you a new target of a
conditional sentence at line 15:
1. |(∃x)[Hx & (∃y)(Dy & Lxy)]
P
2. |(∃x)[Hx & (∀y)(Dy ⊃ ~Lxy)]
P
3. | a |Ha & (∃y)(Dy & Lay)
A
4. | |(∃y)(Dy & Lay)
3 &E
5. | | b |Db & Lab
A
6. | | | c |Hc & (∀y)(Dy ⊃ ~Lcy)
A
7. | | | | |De
A
8. | | | | |(∀y)(Dy ⊃ ~Lcy)
6 &E
9. | | | | |De ⊃ ~Lce
8 ∀E
10. | | | | |~Lce
7,9 ⊃∃
11. | | | | |Lce v ~Lce
10 vI
12. | | | | |Hc
6 &E
13. | | | | |Hc & (Lce v ~Lce)
11,12 &I
14. | | | | |(∃y)[Hy & (Lye v ~Lye)]
13 ∃I
15. | | | |Dê ⊃ (∃y)[Hy & (Lyê v ~Lyê)] 7-14 ⊃I
16. | | | |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]} 15 ∀I
17. | | |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]} 2, 6-16 ∃E
18. | |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]} 4, 5-17 ∃E
19. |(∀x){Dx ⊃ (∃y)[Hy & (Lyx v ~Lyx)]}
1, 3-18 ∃E
Can you think of another way to derive ‘Lce v ~Lce’? Look at
line 13 – can you think of a rule of inference that we could
introduce that would allow us to go directly from line 12 to
line 13? (Hint: remember LTC from our section on Logical
Equivalency.)