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.