Lesson test: Models and countermodels

1.[3p]
What does an interpretation have to supply?

Select all that apply

A non-empty domain

Correct

An object in the domain for each name

Correct

An extension for each predicate

Correct

A truth value for each atom of propositional logic

The answer is: A non-empty domain, An object in the domain for each name, An extension for each predicate

2.[2p]
A countermodel to an argument is an interpretation in which

every premise is true and the conclusion is false

Correct

every premise is false

The answer is: every premise is true and the conclusion is false

the conclusion is false

The answer is: every premise is true and the conclusion is false

no predicate has an extension

The answer is: every premise is true and the conclusion is false

3.[3p]
Why is the domain required to be non-empty?

So that $\forall x F x \to \exists x F x$ comes out valid

Correct

Because predicates need something to be true of

The answer is: So that $\forall x\, Fx \rightarrow \exists x\, Fx$ comes out valid

Because names must refer

The answer is: So that $\forall x\, Fx \rightarrow \exists x\, Fx$ comes out valid

Otherwise no formula could be false

The answer is: So that $\forall x\, Fx \rightarrow \exists x\, Fx$ comes out valid

4.[2p]
With a domain of three objects, how many possible extensions does a one-place predicate have?

CorrectNot quite: 8

5.[3p]
With a domain of two objects, how many possible extensions does a two-place predicate have?

CorrectNot quite: 16

6.[3p]
Which countermodel refutes " $\forall x (F x \to G x)$ , $\exists x G x$ , therefore $\exists x F x$ "?

Domain ${1}$ , $F$ empty, $G$ true of 1

Correct

Domain ${1}$ , $F$ true of 1, $G$ true of 1

The answer is: Domain $\lbrace 1 \rbrace$, $F$ empty, $G$ true of 1

Domain ${1, 2}$ , $F$ true of both, $G$ empty

The answer is: Domain $\lbrace 1 \rbrace$, $F$ empty, $G$ true of 1

Domain ${1}$ , $F$ and $G$ both empty

The answer is: Domain $\lbrace 1 \rbrace$, $F$ empty, $G$ true of 1

7.[2p]
Every satisfiable first-order formula has a finite model.

True

The answer is: False

False

Correct

8.[3p]
Church and Turing showed in 1936 that

no algorithm decides whether an arbitrary first-order formula is valid

Correct

first-order logic is inconsistent

The answer is: no algorithm decides whether an arbitrary first-order formula is valid

validity can be checked by a truth table with enough rows

The answer is: no algorithm decides whether an arbitrary first-order formula is valid

every valid formula has a countermodel

The answer is: no algorithm decides whether an arbitrary first-order formula is valid

9.[2p]
Failing to find a countermodel in domains of one, two and three objects proves that an argument is valid.

True

The answer is: False

False

Correct

1.[3p] What does an interpretation have to supply?

2.[2p] A countermodel to an argument is an interpretation in which

3.[3p] Why is the domain required to be non-empty?

4.[2p] With a domain of three objects, how many possible extensions does a one-place predicate have?

5.[3p] With a domain of two objects, how many possible extensions does a two-place predicate have?

6.[3p] Which countermodel refutes "∀x(Fx→Gx), ∃xGx, therefore ∃xFx"?

7.[2p] Every satisfiable first-order formula has a finite model.

8.[3p] Church and Turing showed in 1936 that

9.[2p] Failing to find a countermodel in domains of one, two and three objects proves that an argument is valid.

1.[3p]
What does an interpretation have to supply?

2.[2p]
A countermodel to an argument is an interpretation in which

3.[3p]
Why is the domain required to be non-empty?

4.[2p]
With a domain of three objects, how many possible extensions does a one-place predicate have?

5.[3p]
With a domain of two objects, how many possible extensions does a two-place predicate have?

6.[3p]
Which countermodel refutes " $\forall x (F x \to G x)$ , $\exists x G x$ , therefore $\exists x F x$ "?

7.[2p]
Every satisfiable first-order formula has a finite model.

8.[3p]
Church and Turing showed in 1936 that

9.[2p]
Failing to find a countermodel in domains of one, two and three objects proves that an argument is valid.