Let D be the set of all students at your school, and let M(s) be "s is a math major," let C(s) be "s is a computer science student," and let E(s) be "s is an engineering student." Express each of the following statements using quantifiers, variables, and the predicates M(s),C(s), and E(s).

a. There is an engineering student who is a math major.

b. Every computer science student is an engineering student.

c. No computer science students are engineering students.

d. Some computer science students are also math majors.

e. Some computer science students are engineering students and some are not.

a. ∃s ∈D such that M(s) ∧ E(s)

b. Vs ∈D, C(s) → E(s)

c. Vs ∈D, C(s) → ~ E(s)

d. ∃s ∈D such that M(s) ∧ C(s)

e. (∃s ∈D such that C(s) ∧ E(s)) ∧ (∃t ∈D such that C(t) ∧ ~ E(t))

Step-by-step explanation:

Given

D = set of all students in your school Doman

M(s) = "is a Math major"

C(s) = "is a computer science student"

E(s) = "is an engineering student"

a. There is an engineering student who is a math major.

We can rewrite the above expression as:

"There's at least one student s ∈ D such that s is a math major and s is an engineering student".

The statement "There's at least one" implies an existential statement.

An existential statement is defined as ∃x:P(x) and it is true if amd only if there is an existence of at least one element in the domain.

So, we'll replace

"s is a math major" with M(s)

"s is an engineering student" with E(s)

"and" with ∧.

Bringing them together, we have

∃s ∈D such that M(s) ∧ E(s)

b. Every computer science student is an engineering student.

We can rewrite the above expression as:

"For every student s ∈ D if s is a computer science student then s is an engineering student".

The statement "For every" implies an universal statement.

A universal statement is defined as VxP(x) and is true if and only if P(x) is true for all values of X in the domain.

So, we'll replace

"s is a computer science student" with C(s)

"s is an engineering student" with E(s)

"If then" with →

Bringing them together, we have

Vs ∈D, C(s) →E(s)

c. No computer science students are engineering students.

We can rewrite the above expression as:

"For every student s ∈ D if s is a computer science student then s is not an engineering student".

The statement "For every" implies an universal statement.

A universal statement is defined as VxP(x) and is true if and only if P(x) is true for all values of X in the domain.

So, we'll replace

"s is a computer science student" with C(s)

"s is an engineering student" with E(s)

"If then" with →

"not" by ~

Bringing them together, we have

Vs ∈D, C(s) → ~ E(s)

d. Some computer science students are also math majors.

We can rewrite the above expression as:

"There's at least one student s ∈ D such that s is a math major and s is an computer science student".

The statement "There's at least one" implies an existential statement.

An existential statement is defined as ∃x:P(x) and it is true if and only if there is an existence of at least one element in the domain.

So, we'll replace

"s is a math major" with M(s)

"s is a computer science student" with C(s)

"and" with ∧.

Bringing them together, we have

∃s ∈D such that M(s) ∧ C(s)

e. Some computer science students are engineering students and some are not.

We can rewrite the above expression as:

"There's at least one student s ∈ D such that s is a computer science student and s is an engineering student" and "There's at least one student t ∈ D such that t is a computer science student and t is not an engineering student"

The statement "There's at least one" implies an existential statement.

An existential statement is defined as ∃x:P(x) and it is true if and only if there is an existence of at least one element in the domain.

So, we'll replace

"s is a computer science student" with C(s)

"s is an engineering student" with E(s)

"and" with ∧.

"not" with ~

Bringing them together, we have

(∃s ∈D such that C(s) ∧ E(s)) ∧ (∃t ∈D such that C(t) ∧ ~ E(t))

hi! i think the answer is 1/2 : )