Translation, Negation, And Witness Identification Quiz

1) Which natural language phrase indicates an existential statement?

For all

Some

None

Always

Existential statements are indicated by phrases that suggest existence, such as "some", "there exists", "at least one", or "a few". "For all" and "always" indicate universal quantification, while "none" indicates the absence of existence.

Explanation

Existential statements are indicated by phrases that suggest existence, such as "some", "there exists", "at least one", or "a few". "For all" and "always" indicate universal quantification, while "none" indicates the absence of existence.

2) A single example can prove an existential statement.

True

False

An existential statement ∃ x P(x) claims that there exists at least one x satisfying P(x). Therefore, providing a single example where P(x) is true is sufficient to prove the statement. This is in contrast to universal statements, which require showing truth for all x.

Explanation

An existential statement ∃ x P(x) claims that there exists at least one x satisfying P(x). Therefore, providing a single example where P(x) is true is sufficient to prove the statement. This is in contrast to universal statements, which require showing truth for all x.

3) Which is equivalent to ∃ x P(x) ∨ ∃ x Q(x)?

∃ x (P(x) ∨ Q(x))

∃ x (P(x) ∧ Q(x))

∀x(P(x) ∨ Q(x))

None of the above

The existential quantifier distributes over disjunction, meaning "there exists an x with P(x) or there exists an x with Q(x)" is equivalent to "there exists an x such that P(x) or Q(x) is true.

Explanation

The existential quantifier distributes over disjunction, meaning "there exists an x with P(x) or there exists an x with Q(x)" is equivalent to "there exists an x such that P(x) or Q(x) is true.

4) ∃ x ∃ y P(x,y) is equivalent to ∃ y ∃ x P(x,y).

True

False

The order of existential quantifiers does not affect the meaning; ∃ x ∃ y P(x,y) means there exist some x and some y such that P(x,y) is true, which is the same as ∃ y ∃ x P(x,y). Both expressions are logically equivalent.

Explanation

The order of existential quantifiers does not affect the meaning; ∃ x ∃ y P(x,y) means there exist some x and some y such that P(x,y) is true, which is the same as ∃ y ∃ x P(x,y). Both expressions are logically equivalent.

5) The statement "There exists a unique x such that P(x)" is written:

∃ x P(x)

∃! x P(x)

∀ x P(x)

¬∃ x P(x)

The symbol for "there exists a unique x" is ∃! x P(x), which means that there is exactly one x satisfying P(x). This notation is standard in logic and distinguishes uniqueness from mere existence.

Explanation

The symbol for "there exists a unique x" is ∃! x P(x), which means that there is exactly one x satisfying P(x). This notation is standard in logic and distinguishes uniqueness from mere existence.

6) From ∃ x ∀ y P(x,y) we can infer:

∀ y ∃ x P(x,y)

∃ x P(x,c) for any c

∀ x ∀ y P(x,y)

Nothing weaker

If there exists an x such that for all y, P(x,y) is true, then for each y, there exists some x (namely, the same x) such that P(x,y) is true. Therefore, we can infer ∀ y ∃ x P(x,y). Note that the converse is not necessarily true.

Explanation

If there exists an x such that for all y, P(x,y) is true, then for each y, there exists some x (namely, the same x) such that P(x,y) is true. Therefore, we can infer ∀ y ∃ x P(x,y). Note that the converse is not necessarily true.

7) The minimal domain size to make ∃ x ∃ y (x ≠ y) true is:

0

1

2

Infinite

The statement ∃ x ∃ y (x ≠ y) requires that there exist two distinct elements x and y in the domain. Therefore, the domain must have at least two elements.

Explanation

The statement ∃ x ∃ y (x ≠ y) requires that there exist two distinct elements x and y in the domain. Therefore, the domain must have at least two elements.

8) ∃ x P(x) ∧ ∃ x ¬P(x) can be true in the same model.

True

False

This conjunction means that there exists some x with P(x) true and there exists some x with P(x) false. This can be true if the domain has at least two elements.

Explanation

This conjunction means that there exists some x with P(x) true and there exists some x with P(x) false. This can be true if the domain has at least two elements.

9) Which is equivalent to ∃ x P(x) ∨ ∀ x ¬P(x)?

Tautology

Contradiction

Contingent

¬∃ x P(x)

This disjunction is always true because either there exists some x with P(x) true, or for all x, P(x) is false (which is equivalent to ¬∃ x P(x)). Thus, it covers all possible cases and is a tautology.

Explanation

This disjunction is always true because either there exists some x with P(x) true, or for all x, P(x) is false (which is equivalent to ¬∃ x P(x)). Thus, it covers all possible cases and is a tautology.

10) From ∀ x (P(x) → Q(x)) and ∃ x ¬Q(x) we can infer:

∃x ¬P(x) ∧ ∃x Q(x)

∀ x ¬P(x)

∃ x ¬P(x)

Nothing

From ∃x ¬Q(x), there exists some c with ¬Q(c). From ∀x(P(x) → Q(x)), we have P(c) → Q(c). Since ¬Q(c) is true, P(c) must be false by modus tollens, so ¬P(c) is true, and therefore ∃x ¬P(x).

Explanation

From ∃x ¬Q(x), there exists some c with ¬Q(c). From ∀x(P(x) → Q(x)), we have P(c) → Q(c). Since ¬Q(c) is true, P(c) must be false by modus tollens, so ¬P(c) is true, and therefore ∃x ¬P(x).

11) The difference between ∃ x ∀ y P(x,y) and ∀ y ∃ x P(x,y) is:

None, they are equivalent

∃ x ∀ y is stronger

∀ y ∃ x is stronger

They are unrelated

∃ x ∀ y P(x,y) means there is a single x that works for all y, while ∀ y ∃ x P(x,y) means that for each y, there is some x that may depend on y. Therefore, ∃ x ∀ y P(x,y) implies ∀ y ∃ x P(x,y), but not vice versa, making ∃ x ∀ y stronger.

Explanation

∃ x ∀ y P(x,y) means there is a single x that works for all y, while ∀ y ∃ x P(x,y) means that for each y, there is some x that may depend on y. Therefore, ∃ x ∀ y P(x,y) implies ∀ y ∃ x P(x,y), but not vice versa, making ∃ x ∀ y stronger.

12) Which is a correct inference rule for ∃?

∃-introduction: from P(c) infer ∃ x P(x)

∃-elimination: from ∃ x P(x) infer P(c)

∃ distributes over ∧

¬∃ x P(x) ≡ ∃ x ¬P(x)

Existential introduction is a valid inference rule: if P(c) is true for some constant c, then we can infer that there exists some x such that P(x). Existential elimination does not allow inferring P(c) for arbitrary c; it requires a subproof. ∃ does not distribute over ∧, and ¬∃ x P(x) is equivalent to ∀ x ¬P(x), not ∃ x ¬P(x).

Explanation

Existential introduction is a valid inference rule: if P(c) is true for some constant c, then we can infer that there exists some x such that P(x). Existential elimination does not allow inferring P(c) for arbitrary c; it requires a subproof. ∃ does not distribute over ∧, and ¬∃ x P(x) is equivalent to ∀ x ¬P(x), not ∃ x ¬P(x).

13) True or False: ∃ x (P(x) ∧ Q(x)) implies ∃ x P(x) ∧ ∃ x Q(x).

True

False

If there exists an x such that both P(x) and Q(x) are true, then certainly there exists an x such that P(x) is true, and there exists an x such that Q(x) is true. Therefore, the implication holds.

Explanation

If there exists an x such that both P(x) and Q(x) are true, then certainly there exists an x such that P(x) is true, and there exists an x such that Q(x) is true. Therefore, the implication holds.

14) “∃ x ∀ y (x > y)” over real numbers is:

True

False

Undefined

Contradiction

Over the real numbers, there is no number x that is greater than all other numbers y. For any x, there exists a y such that y > x (e.g., y = x + 1). Therefore, this statement is false.

Explanation

Over the real numbers, there is no number x that is greater than all other numbers y. For any x, there exists a y such that y > x (e.g., y = x + 1). Therefore, this statement is false.

15) The negation of ∃ x ∀ y P(x,y) is:

∀ x ∃ y ¬P(x,y)

∀ x ∀ y ¬P(x,y)

∃ x ∃ y ¬P(x,y)

¬∃ x ∀ y P(x,y)

The negation of ∃ x ∀ y P(x,y) is ¬∃ x ∀ y P(x,y), which is equivalent to ∀ x ¬∀ y P(x,y), and further equivalent to ∀ x ∃ y ¬P(x,y). This means that for every x, there is some y for which P(x,y) is false.

Explanation

The negation of ∃ x ∀ y P(x,y) is ¬∃ x ∀ y P(x,y), which is equivalent to ∀ x ¬∀ y P(x,y), and further equivalent to ∀ x ∃ y ¬P(x,y). This means that for every x, there is some y for which P(x,y) is false.

Translation, Negation, and Witness Identification Quiz