Lesson test: Identity, number and descriptions

1.[3p]
Why does $\exists x F x \land \exists y F y$ fail to say that there are two $F$ s?

Two variables may take the same value, so one object satisfies it

Correct

It has a free variable

The answer is: Two variables may take the same value, so one object satisfies it

The conjunction should be a disjunction

The answer is: Two variables may take the same value, so one object satisfies it

It says only that something is $F$ or something is $G$

The answer is: Two variables may take the same value, so one object satisfies it

2.[3p]
"At most one $F$ " is written as

$\forall x \forall y ((F x \land F y) \to x = y)$

Correct

$\exists x \forall y (F y \to y = x)$

The answer is: $\forall x \forall y ((Fx \wedge Fy) \rightarrow x = y)$

$\exists x \exists y (F x \land F y \land x \neq y)$

The answer is: $\forall x \forall y ((Fx \wedge Fy) \rightarrow x = y)$

$\forall x (F x \to x = x)$

The answer is: $\forall x \forall y ((Fx \wedge Fy) \rightarrow x = y)$

3.[2p]
$\exists x (F x \land \forall y (F y \to y = x))$ says

there is exactly one $F$

Correct

there is at most one $F$

The answer is: there is exactly one $F$

there are at least two $F$ s

The answer is: there is exactly one $F$

everything is $F$

The answer is: there is exactly one $F$

4.[2p]
"At most one $F$ " is true when there are no $F$ s at all.

True

Correct

False

The answer is: True

5.[3p]
On Russell's analysis, "the present King of France is bald" is

false, because nothing satisfies its existence clause

Correct

neither true nor false

The answer is: false, because nothing satisfies its existence clause

true, vacuously

The answer is: false, because nothing satisfies its existence clause

meaningless

The answer is: false, because nothing satisfies its existence clause

6.[3p]
Which parts does Russell's analysis of "the $F$ is $G$ " contain?

Select all that apply

Something is $F$

Correct

At most one thing is $F$

Correct

That thing is $G$

Correct

Everything is $G$

The answer is: Something is $F$, At most one thing is $F$, That thing is $G$

7.[2p]
Why does identity count as a logical symbol rather than ordinary vocabulary?

Its extension is fixed in every interpretation

Correct

It takes two arguments

The answer is: Its extension is fixed in every interpretation

It appears in every formula

The answer is: Its extension is fixed in every interpretation

It cannot be negated

The answer is: Its extension is fixed in every interpretation

8.[2p]
First-order logic with identity can express "there are finitely many $F$ s".

True

The answer is: False

False

Correct

9.[3p]
Which theorem shows that no first-order sentence is true in exactly the finite domains?

Compactness

Correct

Soundness

The answer is: Compactness

Excluded middle

The answer is: Compactness

Leibniz's law

The answer is: Compactness

1.[3p] Why does ∃xFx∧∃yFy fail to say that there are two Fs?

2.[3p] "At most one F" is written as

3.[2p] ∃x(Fx∧∀y(Fy→y=x)) says

4.[2p] "At most one F" is true when there are no Fs at all.

5.[3p] On Russell's analysis, "the present King of France is bald" is

6.[3p] Which parts does Russell's analysis of "the F is G" contain?

7.[2p] Why does identity count as a logical symbol rather than ordinary vocabulary?

8.[2p] First-order logic with identity can express "there are finitely many Fs".

9.[3p] Which theorem shows that no first-order sentence is true in exactly the finite domains?

1.[3p]
Why does $\exists x F x \land \exists y F y$ fail to say that there are two $F$ s?

2.[3p]
"At most one $F$ " is written as

3.[2p]
$\exists x (F x \land \forall y (F y \to y = x))$ says

4.[2p]
"At most one $F$ " is true when there are no $F$ s at all.

5.[3p]
On Russell's analysis, "the present King of France is bald" is

6.[3p]
Which parts does Russell's analysis of "the $F$ is $G$ " contain?

7.[2p]
Why does identity count as a logical symbol rather than ordinary vocabulary?

8.[2p]
First-order logic with identity can express "there are finitely many $F$ s".

9.[3p]
Which theorem shows that no first-order sentence is true in exactly the finite domains?