Question

PREDICATE LOGIC:

A.Write the following predicates symbolically and determine their true value.

Note: Use at least three (3) values for the variables.

1. for every real number x, if x>1 then x – 1 > 1

2. for some real number x, x2 ≤ 0 

B. Translate the following English sentence into a symbol. (3 pts each)

1. No one in this class is wearing pants and a guitarist.

Let:

Domain of x is all persons

A(x): x is wearing pants

B(x): x is a guitarist

C(x): belongs to the class

Answer:

2. No one in this class is wearing pants and a guitarist.

Let:

The domain of x is persons in this class

A(x): x is wearing pants

B(x): x is a guitarist

Answer:

3. There is a student at your school who knows C++ but who doesn’t

know Java.

Let:

Domain: all students at your school

C(x): x knows C++

J(x): x knows Java

Answer:

Answer 1

PREDICATE LOGIC:

A.Write the following predicates symbolically and determine their true value.

Note: Use at least three (3) values for the variables.

1. for every real number x, if x>1 then x – 1 > 1

Answer:

Let:

Domain of x is all real number

P(x) : x > 1

Q(x) : x – 1 > 1

Answer: $\forall$x(P(x) → Q(x))

Truth value: x = {3,4,5} satisfies this statement

2. for some real number x, $x^2 \le 0$

Answer:

Domain of x is all real number

P(x) : $x^2 \le 0$

Answer: $\exist$x(P(x))

Truth value: there is only one value x = 0 satisfies predicate P(x)

B. Translate the following English sentence into a symbol. (3 pts each)

1. No one in this class is wearing pants and a guitarist.

Let:

Domain of x is all persons

A(x): x is wearing pants

B(x): x is a guitarist

C(x): belongs to the class

Answer: !$\exist$x(C(x) and A(x) and B(x))

2. No one in this class is wearing pants and a guitarist.

Let:

The domain of x is persons in this class

A(x): x is wearing pants

B(x): x is a guitarist

Answer: !$\exist$x(A(x) and B(x))

3. There is a student at your school who knows C++ but who doesn’t know Java.

Let:

Domain: all students at your school

C(x): x knows C++

J(x): x knows Java

Answer: $\exist$x(C(x) and !J(x))