Formal Logic Assignment: Truth Values, Predicate Logic, and WFFs

This assignment delves into the core concepts of formal logic, presenting solutions to problems involving truth values within specific interpretations and domains. The assignment explores the evaluation of well-formed formulas (wffs) using both integer and state-based domains, determining their truth or falsity. It also tackles predicate logic, including the use of quantifiers (∀ and ∃) to represent statements about the world, such as relationships between states and the properties of objects. Furthermore, the assignment covers the translation of English language statements into predicate wffs and vice-versa, enabling a deeper understanding of logical equivalency and the construction of logical arguments. The solutions include negations of statements, demonstrating a comprehensive understanding of the subject matter.

Assignment 2.1 - Section 1.3 Exercise
2. What is the truth value of each of the following wffs in the interpretation where the
domain consists of the integers?
a. (4x)(E y)(x + y = x) e. (4x)(4y)(x < y ~ y < x)
b. (E y)(4x)(x + y = x) f. (4x)[x < 0 S (E y)` x + y = 0)
c. (4x)(E y)(x + y = 0) g. (E x)(E y)(x2 = y)
d. (E y)(4x)(x + y = 0) h. (4x)(x2 > 0)
Solution
a. (E y)(x + y = x)
b. (4x)’(x + y = x)
c. (E y)(x + y = 0)
d. (4x)’(x + y = 0)
e. (4y)(x <y )
f. (E y)( x + y = 0)
g. (E x)(x2 = y)
h. (E x)(x2 > 0)
3. Give the truth value of each of the following wffs in the interpretation where the
domain consists ofthe states of the United States, Q(x, y) is “x is north of y,” P(x) is “x
starts with the letter M,” and a is
“Massachusetts.”
a.(4x)P(x)
Solution
i. (∀x)P(x) → (∃x)P(x
ii. (∀x)P(x) → P(a)
b. (4x)(4y)(4z)[Q(x, y) ` Q( y, z) S Q(x, z)]
Solution
(4x)[( Q(x, y) → ` Q( y, z) → S Q(x, z)]
c. (E y)(E x)Q( y, x)
Solution
(E x)Q( y, x)
i. (4x)(E y)[P( y) ` Q(x, y)]
(E y)[P( y)’ → Q(x, y)]
j. (E y)Q(a, y)
Solution
(E y) → Q(a, y)
f. (E x)[P(x) ` Q(x, a)]

Solution
P(x) ` → Q(x, a)
12. Which of the following sentences are equivalent to the statement
Cats are smarter than dogs.
a. Some cats are smarter than some dogs.
Solution
(∃x)[C(x) ∧ (∃y)(D( y) → S(x, y))]
b. There is a cat that is smarter than all dogs.
Solution
C(x)→(∀y)(D(y) → S(x, y)
c. All cats are smarter than all dogs.
Solution
(∀x)[C(x)→(∀y)(D(y) → S(x, y))]
d. Only cats are smarter than dogs.
Solution
(∀x)(C(x)→(∀y)[(S(y) → D(x, y))]
e. All cats are smarter than any dog.
Solution
(∀x)C(x)→(D(y) → S(x, y))
15. Using the predicate symbols shown and appropriate quantifiers, write each English
language statement as
a predicate wff. (The domain is the whole world.)
M(x): x is a man
W(x): x is a woman
T(x): x is tall
a. All men are tall.
Solution
(∀x)[(M(x)→(T(x)]
b. Some women are tall.
Solution
(∃x)[(W(x)→(T(x)]
c. All men are tall but no woman is tall.

Solution
(∀x)[M(x)→[(∀x)(W(x)]’ → T(x))]
d. Only women are tall
Solution
(∀x)[(W(x)→( ∀x)(T(x)→(T(x)]
e. No man is tall.
Solution
[(∀x)[(M(x)]’→(T(x)
f. If every man is tall, then every woman is tall.
Solution
[M(x)→(W(x) → T(x))]
g. Some woman is not tall.
Solution
[(∃x)[(W(x)→(T(x)]]’
h. If no man is tall, then some woman is not tall.
Solution
[[(∀x)[M(x)]’→[(∀x)(W(x) → T(x))]]
25. Give English language translations of the following wffs if
L(x, y): x loves y
H(x): x is handsome
M(x): x is a man
P(x): x is pretty
W(x): x is a woman
j: John
k: Kathy
L(x, y): x loves y
a. H( j) ` L(k, j)
Kathy loves John because he is handsome.
b. (4x)[M(x) S H(x)]
All men are handsome.
c. (4x)(W(x) S (4y)[L(x, y) S M( y) ` H( y)])
All women love all handsome men.
d. (E x)[M(x) ` H(x) ` L(x, k)]
Kathy does not love some men who are not handsome.
e. (E x)(W(x) ` P(x) ` (4y)[L(x, y) S H( y) ` M( y)])
Some pretty women do not alllove men who are not handsome.

f. (4x)[W(x) ` P(x) S L( j, x)]
Not all pretty women love John.
26. Give English language translations of the following wffs if
M(x): x is a man
W(x): x is a woman
i: Ivan
p: Peter
W(x, y): x works for y
a. (E x)(W(x) ` (4y)(M( y) S [W(x, y)]′))
Solution
Some women do not work for only all men.
b. (4x)[M(x) S (E y)(W( y) ` W(x, y))]
Solution
No woman do some work for all men.
c. (4x)[M(x) S (4y)(W(x, y) S W( y))]
Solution
All women work for all men.
56 Formal Logic
d. (4x)(4y)(M(x) ` W( y, x) S W( y))
Solution
All women do not work for all men.
e. W(i, p) ` (4x)[W( p, x) S (W(x))′])
Solution
All women do not work for Ivans or Peter
f. (4x)[W(x, i) S (W(x))′]
solution
All women do not work for Ivans.
31. Write the negation of each of the following statements.
a. Some farmer grows only corn.
Solution
Some farmer do not grow only corn.
g. All farmers grow corn.
Solution
Not all farmers grows corn.
h. Corn is grown only by farmers.
Solution
Corn is not grown only by farmers.
32. Write the negation of each of the following statements
a. Some child fears all clowns.
Solution

Some child do not fears all clowns.
b. Some children fear only clowns.
Solution
Some children do not fear only clowns.
c. No clown fears any child.
Solution
All clown fears any child.