Let the domain of x be the set of geometric figures in the plane, and let Square (x) be "x is a square" and Rect (x) be "x is a rectangle." a. ∃x such that Rect (x) ∧ Square (x). b. ∃x such that Rect (x) ∧ ∼Square (x). c. ∀x, Square (x) → Rect (x).

a) The statement is also true, because a square is a geometric figure in the plane and a square is also a rectangle.

b) The statement is also true, because for example a rectangle with length 4 and width 6 is a possible rectangle that is not a square (as the length and the width are unequal).

c) The statement is true, because squares are a specific type of rectangle and thus all squares are rectangles.

Step-by-step explanation:

Negation ~P: not P

Conjunction p ∧ q: p and q

Conditional statement p - -> q: if p, then q

Existential statement ∃xP (x) is true if and only one element x in the domain for which P (x) is true.

Universal statement ∀xP (x) is true if and only if P (x) is true for all values of x in the domain.

Domain=set of all geometric figures in the plane

Square (x) = "x is a square"

Rect (x) = "x is a rectangle"

(a)

∃x such that Rect (x) ∧ Square (x)

∃ mean "there exists"

In words, the given statement then means that: There exists a geometric figure in the plane that is a rectangle and that is a square.

The statement is also true, because a square is a geometric figure in the plane and a square is also a rectangle.

(b)

∃x such that Rect (x) ∧ ~ Square (x)

∃ mean "there exists"

In words, the given statement then means that: There exists a geometric figure in the plane that is a rectangle and that is not a square.

The statement is also true, because for example a rectangle with length 4 and width 6 is a possible rectangle that is not a square (as the length and the width are unequal).

(c)

∀x, Square (x) - > Rect (x)

∀ means "for every".

In words, the given statement means that: All squares are also rectangles.

The statement is true, because squares are a specific type of rectangle and thus all squares are rectangles.