(~ p  q)  (~ r  ~ q)
s  (r  p)
~ (s  r)
s
r  p
~ p  q
~ r  ~ q
(8)  1.
 2.
 3.
4.
5.
6.
7.
8.
~s
~r
r
9.
From line 1
From line 3
p
~p
p
p
Negation of conclusion
From line 2
From line 4
q
≁≁r
From line 5
~q
From line 5
All paths closed. Valid.
VII. CHAPTER SEVEN: PREDICATE LOGIC SYMBOLIZATION
A. General theory:
1. Which of the following are sentences?
a. (x)Fx  (y)Gxy
b. (x)Fx  Ga
c. (x)(Fx  Ga)
d.
e.
~ (y) Fy  (x)Gx
None of these
2. Which of the following are sentences?
a. (x)(Fx  Gx)
b. (x)Fx  (x)Gx
c. (x)Fx  Gx
d.
e.
(x)Fy  (y)Gx
None of these
3. Symbolize “Some mammals are not four-legged” a) when the domain is mammals
and b) when the domain is unrestricted (using obvious abbreviations).
4. Symbolize “No whales are fish” a) when the domain is whales, and b) when the
domain is unrestricted (using obvious abbreviations).
5. What sentence contradicts “No just acts are acts that cause pain”?
a. All just acts are acts that cause pain.
b. Some just acts are acts that cause pain.
c. Some just acts are acts that do not cause pain.
d. No acts that cause pain are just acts.
e. None of these.
23
6. Which sentence is equivalent to (x)(Ax ∙ ~ Bx)?
a. (x)(Ax  ~ Bx)
c. ~ (x)(Ax  Bx)
b. (x)(Ax  Bx)
d. None of these
7. Which sentence is equivalent to (x)(Ax ∙ Bx)?
a. ~ (x)(Ax  ~ Bx)
c.
b. ~ (x)(Ax ∙ ~ Bx)
d.
~ (x)(Ax  Bx)
None of these
A. Answers
1.
2.
3.
4.
5.
6.
7.
b, c, d
a, b
a) (x) ~ Fx, b) (x)(Mx ∙ ~ Fx)
a) ~ (x)Fx or (x) ~ Fx, b) ~ (x)(Wx ∙ Fx) or (x)(Wx  ~ Fx)
b
c
a
B. Symbolize, using the indicated letters:
1.
No mathematician philosophers are scientists. (Mx = “x is a mathematician”; Px =
“x is a philosopher”; Sx = “x is a scientist”.)
2.
All mathematicians and philosophers are either nonphilosophers or
nonscientists.
3.
Some mathematicians and (some) philosophers are scientists.
4.
No one is a scientist unless he also is a mathematician and philosopher.
5.
Only mathematicians are scientists, and none but philosophers are
mathematicians.
6.
No arguments that are either invalid or unsound are convincing. (Ax = “x is an
argument”, Vx = “x is valid”; Sx = “x is sound”; Cx = “x is convincing”.)
7.
A student caught cheating will be expelled. (Sx = “x is a student”; Cx = “x is
caught cheating”; Ex = “x will be expelled”.)
8.
A student who was caught cheating was not expelled.
9.
If all students are good logicians, then they all will pass. (Sx = “x is a student”; Gx
= “x is a good logician”; Px = “x will pass”.)
10.
If a student is a good logician, then he or she will pass.
24
11.
If not all students will pass, then some students are not good logicians.
12.
If any students pass, then they studied hard. (Hx = “x studied hard”.)
13.
If all students pass, then at least some of them are good logicians.
14.
Some students are good logicians, but are such that they won’t pass unless they
study hard.
15.
Some students are neither good logicians nor good writers, but some students are
both. (Wx = “x is a good writer”.)
B. Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
(x)[(Mx  Px)  ~ Sx] or ~ (x)[(Mx  Px)  Sx]
(x)[(Mx  Px)  (~ Px  ~ Sx)]
(x)(Mx  Sx)  (x)(Px ∙ Sx)
(x)[Sx  (Mx  Px)] or ~ (x)[Sx  ~ (Mx  Px)]
(x)(Sx  Mx) ∙ (x)(Mx  Px)
(x){[Ax  (~ Vx  ~ Sx)]  ~ Cx} or ~ (x){[Ax  (~ Vx  ~ Sx)]  Cx}
(x)[(Sx  Cx)  Ex]
(x)[(Sx ∙ Cx) ∙ ~ Ex]
(x)(Sx  Gx)  (x)(Sx  Px)
(x)[(Sx ∙ Gx)  Px]
~ (x)(Sx  Px)  (x)(Sx  ~ Gx)
(x)[(Sx  Px)  Hx]
(x)(Sx  Px)  (x)(Sx  Gx)
(x)[(Sx ∙ Gx) ∙ (~ Hx  ~ Px)] or (x)[(Sx ∙ Gx) ∙ (Px  Hx)]
(x)[Sx ∙ ~ (Gx  Wx)] ∙ (x)[Sx ∙ (Gx ∙ Wx)]
C. Translations. Translate the following into English, being a colloquial as possible. Use
same symbols as above, i.e., let Sx = “x is a student”; Cx = “x is caught cheating”;
Ex = “x will be expelled”; Gx = “x is a good logician”; Px = “x will pass”; Hx =
“x studied hard”; Wx = “x is a good writer”; and let Ax = “x is angry”; Yx = “x is
happy”; a = Alice; b = Burt.
1.
2.
3.
4.
5.
6.
7.
(x)(Sx ∙ Cx)  Aa
(Wb ∙ Wa) ∙ (Ga ∙ ~ Gb)
(x)[(Sx ∙ Hx)  Px] ∙ (x) [(Sx ∙ Hx) ∙ ~ Yx]
(x)[(Sx ∙ Yx)  ~ Wx] ∙ ~ (x)[ [(Sx ∙ Wx)  Ax]
(x)(Sx  Hx)  (x)(Sx  ~ Cx)
(x)[(Sx ∙ Hx)  ~ Cx]
(x){(Sx ∙ Ax)  [Wx ∙ ~ (Hx  Gx)]}
25
C. Answers
1.
2.
3.
4.
5.
6.
7.
If a student is caught cheating then Alice is angry.
Burt and Alice are good writers, and Alice is a good logician but Burt is not.
All students who study hard will pass, but some are not happy.
No happy students are good writers, but not all students who are good writers are
angry.
If all students studied hard, then none will be caught cheating.
No students who studied hard are caught cheating.
An angry student will be a good writer but will neither study hard nor be good
logician.
VIII. CHAPTER EIGHT: PREDICATE LOGIC SEMANTICS
A. General Theory
1. If there are no unicorns, what is the truth value of the sentence (x)(Ux  Mx), where
Ux = x is a unicorn, and Mx = x is mortal?
2. What about the truth value of the sentence (x)(Ux  Mx)?
A. Answers
1. The sentence (x)(Ux  Mx) will be true. To see this, consider the expansion of this
sentence for larger and larger universes of discourse. First, in a twoindividual universe
of discourse, its expansion is (Ua  Ma)  (Ub  Mb), which will be true since each
conditional will have a false antecedent. Then for larger universes, each new conjunction
added will have a false antecedent and thus be true (on the assumption nothing is a
unicorn).
2. The sentence (x) (Ux  Mx) will be true also. Again, consider its expansion for a
twoindividual universe, namely (Ua  Ma)  (Ub  Mb), which will be true because
each conditional will have a false antecedent. And obviously, it will be true in larger
universes, since adding more disjunctions to a true sentence can't make it false.
B. Proving invalidity. Prove that the following arguments are invalid by either the
interpretation method or by the expansion method:
(1)
1. (x)(Ax  Bx)
2. (x) ~ Ax
/(x) ~ Bx
(3) 1. (x)(~ Ax  Bx)
2. ~ (x) ~ Ax
/~ (x)Bx
(2)
1. (x)[(Ax  Bx)  Cx]
2. ~ (Ba  Cb)
3. ~ (Ca  Ab)
/~ (~ Aa  ~ Ba)
(4) 1.
2.
26
~ (x)(~ Ax  Bx)
(x)(Ax  Cx)
/~ (x)(Cx  ~ Bx)
(5)
1. (x)[(Ax ∙ Bx)  Cx]
/(x)[(Ax  Bx)  Cx]
B. Answers
(1)
Let domain be unrestricted, Ax = x is tall, and Bx = x is identical with itself.
(2)
In a twoindividual universe of discourse, the argument amounts to
1. [(Aa  Ba)  Ca]  [(Ab  Bb)  Cb]
2. ~ (Ba  Cb)
3. ~ (Ca  Ab) / ~ (~ Aa  ~ Ba)
Let Aa, Ba, Ca, Ab, Bb, and Cb all be false; the premises are true, the conclusion
false.
(3)
Let the domain of discourse be restricted to the positive integers, and let Ax = x is
an integer, Bx = x is odd. Then we get
1. All positive integers that are not integers are odd. (True, antecedent always F)
2. It is not the case that there is a positive integer that is not an integer. (True)
/ It is not the case that some positive integer is odd. (False)
(4)
Restrict the domain of discourse to two individuals, a and b. Then the expansion
for the argument in question will be
1. ~ [(~ Aa  Ba)  (~ Ab  Bb)]
2. (Aa  Ca)  (Ab  Cb) / ~ [(Ca  ~ Ba)  (Cb  ~ Bb)]
This argument has true premises and a false conclusion if Aa and Ca are true and
Ba, Bb, Ab, and Cb are false.
(5)
Let the domain of discourse be restricted to the positive integers, and let Ax = x >
10; Bx = x > 5; and Cx = x > 7. Then we get
1. (x){[(x > 10)  (x > 5)]  (x > 7)}
/ (x){[(x > 10)  (x > 5)]  (x > 7)}
which has a true premise and a false conclusion.
C. Proving consistency. Show that the premises of the arguments below are consistent.
(1)
1. (x)[(Ax  Bx)  Cx]
2. (y)(Ay  ~ Cy)
3. (z)(Bz  ~ Cz)
/(x)(~ Ax  ~ Bx)
(2)
1. (x)[(Ax  Bx)  Cx]
2. Aa  Ba
3. ~ Cb /~ Bb
27
C. Answers
(1) Let the domain be the positive integers, Ax = x > 1, Bx = x is odd, Cx = x is > 2:
1. All positive integers that are greater than one and odd are greater than 2.
2. Some positive integer greater than one is not greater than two.
3. Some odd positive integer is not greater than two.
(2) Let the domain and predicates be as in the above example, and let a = 3, b = 2:
1. All positive integers that are greater than one and odd are greater than 2.
2. Three is greater than one and odd.
3. Two is not greater than two.
IX. CHAPTER NINE: PREDICATE LOGIC PROOFS. Prove valid:
(1)
1. ~ (x) Ax
/ (x)(Ax  Bx)
(5)
1. (x)[Ax  (y)By]
/ (x)Ax  (y)By
(2)
1. ~ Aa
/~ (x) (Ax  Bx)
(6)
1. (x)Ax  (y)By
/ (x) [Ax  (y)By]
(3)
1. (x) (Ax  Bx)
2. (x) [Ax  (Cx  Dx)]
/ (x) (Cx  Dx)
(7)
1. (x)[(Ax  Bx)  Cx]
2. (y) ~ (~ Ay  Cy)
/(x)(Ax  Bx)
(4)
1. ~ (x) ~ (~ Ax  Bx)
2. ~ (x)Bx /~ (x)Ax
(8)
1. (x)(Ax  Bx)  ~ (y) Cy
2. ~ (x) ~ Cx
/(x)(Cx  Ax)
IX. ANSWERS
(1)
1.
2.
3.
4.
5.
6.
~ (x) Ax
(x)~ Ax
~ Ax
~ Ax  Bx
Ax  Bx
(x)(Ax  Bx)
p
1 QN
2 UI
3 Add
4 Impl
5 UG
(2)
1. ~ Aa
2. ~ Aa  ~ Ba
3. ~ (Aa  Ba)
4. (x) ~ (Ax  Bx)
5. ~ (x)(Ax  Bx)
p
1 Add
2 DeM
3 EG
4 QN
(3)
28
1.
2.
3.
4.
5.
6.
7.
(x)(Ax  Bx)
(x)[Ax  (Cx  Dx)]
Ax  Bx
Ax
Ax  (Cx  Dx)
Cx  Dx
(x)(Cx  Dx)
p
p
1 EI
3 Simp
2 UI
4,5 MP
6 EG
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
~ (x) ~ (~ Ax  Bx) p
~ (x)Bx
p
~ ~ (x)Ax
AP/~ (x)Ax
(x) ~ Bx
2 QN
~ Bx
4 EI
(x)Ax
3 DN
Ax
6 UI
(x)(~ Ax  Bx)
1 QN
~ Ax  Bx
9 UI
~ ~ Ax
7 DN
Bx
9,10 DS
Bx  ~ Bx
5,11Conj
~ (x)Ax
3-12 IP
(5)
1.
2.
3.
4.
5.
6.
(x)[Ax  (y)By]
p
(x)Ax
AP/(y)By
Ax  (y)By
1 EI
Ax
2 UI
(y)By
3,4 MP
(x)Ax  (y)By
2-5 CP
(6)
1.
2.
3.
4.
5.
6.
7.
8.
(x)Ax  (y)By
~ (y)By
~ (x)Ax
(x) ~ Ax
~ Ax
~ (y)By  ~ Ax
Ax  (y)By
(x)[Ax  (y)By]
(7)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
(4)
(8)
p
AP/ ~ Ax
1,2 MT
3 QN
4 EI
2-5 CP
6 Contra
7 EG
(x)[(Ax  Bx)  Cx] p
(y) ~ (~ Ay  Cy)
p
~ (x)(Ax ∙ Bx)
AP
~ (~ Ay  Cy)
2 EI
~ ~ Ay  ~ Cy
4 DeM
Ay  ~ Cy
5 DN
(Ay  By)  Cy
1 UI
Ay
6 Simp
Ay  By
9 Add
Cy
7, 9 MP
~ Cy
6 Simp
Cy  ~ Cy
10,11 Conj
(x)(Ax ∙ Bx)
3-12 IP
29
1. (x)(Ax  Bx) 
~ (y)Cy
2. ~ (x) ~ Cx
3. (x)Cx
4. Cx
5. (y) Cy
6. ~ ~ (y) Cy
7. ~ (x)(Ax  Bx)
8. (x) ~ (Ax  Bx)
9. ~ (Ay  By)
10. ~ (~ Ay  By)
11. ~ ~Ay  ~ By)
12. ~ ~Ay
13. Ay
14. Ay  ~ Cy
15. ~ Cy  Ay
16. Cy  Ay
17. (x)(Cx  Ax)
p
p
2 QN
3 EI
4 EG
5 DN
1,6 MT
7 QN
8 EI
9 Impl
10 DeM
11 Simp
12 DN
13 Add
14 Comm
15 Impl
16 EG