# Chapter 33 Using interpretations

Suppose we want to show that ‘$\exists x\,A(x,x)\rightarrow B(d)$’ is not a validity. This requires showing that the sentence is not true in every interpretation; i.e., that it is false in some interpretation. If we can provide just one interpretation in which the sentence is false, then we will have shown that the sentence is not a validity.

In order for ‘$\exists x\,A(x,x)\rightarrow B(d)$’ to be false, the antecedent (‘$\exists x\,A(x,x)$’) must be true, and the consequent (‘$B(d)$’) must be false. To construct such an interpretation, we start by specifying a domain. Keeping the domain small makes it easier to specify what the predicates will be true of, so we will start with a domain that has just one member. For concreteness, let’s say it is just the city of Paris.

domain:

Paris

The name ‘$d$’ must refer to something in the domain, so we have no option but:

$d$:

Paris

Recall that we want ‘$\exists x\,A(x,x)$’ to be true, so we want all members of the domain to be paired with themselves in the extension of ‘$A$’. We can just offer:

$A(x,y)$:

blank${}_{x}$ is identical with blank${}_{y}$

Now ‘$A(d,d)$’ is true, so it is surely true that ‘$\exists x\,A(x,x)$’. Next, we want ‘$B(d)$’ to be false, so the referent of ‘$d$’ must not be in the extension of ‘$B$’. We might simply offer:

$B(x)$:

blank${}_{x}$ is in Germany

Now we have an interpretation where ‘$\exists x\,A(x,x)$’ is true, but where ‘$B(d)$’ is false. So there is an interpretation where ‘$\exists x\,A(x,x)\rightarrow B(d)$’ is false. So ‘$\exists x\,A(x,x)\rightarrow B(d)$’ is not a validity.

We can just as easily show that ‘$\exists x\,A(x,x)\rightarrow B(d)$’ is not a contradiction. We need only specify an interpretation in which ‘$\exists x\,A(x,x)\rightarrow B(d)$’ is true; i.e., an interpretation in which either ‘$\exists x\,A(x,x)$’ is false or ‘$B(d)$’ is true. Here is one:

domain:

Paris

$d$:

Paris

$A(x,y)$:

blank${}_{x}$ is identical with blank${}_{y}$

$B(x)$:

blank${}_{x}$ is in France

This shows that there is an interpretation where ‘$\exists x\,A(x,x)\rightarrow B(d)$’ is true. So ‘$\exists x\,A(x,x)\rightarrow B(d)$’ is not a contradiction.

• To show that $\mathscr{A}$ is not a validity, it suffices to find an interpretation where $\mathscr{A}$ is false.

• To show that $\mathscr{A}$ is not a contradiction, it suffices to find an interpretation where $\mathscr{A}$ is true.

## 33.2 Logical equivalence

Suppose we want to show that ‘$\forall x\,S(x)$’ and ‘$\exists x\,S(x)$’ are not logically equivalent. We need to construct an interpretation in which the two sentences have different truth values; we want one of them to be true and the other to be false. We start by specifying a domain. Again, we make the domain small so that we can specify extensions easily. In this case, we will need at least two objects. (If we chose a domain with only one member, the two sentences would end up with the same truth value. In order to see why, try constructing some partial interpretations with one-member domains.) For concreteness, let’s take:

domain:

Ornette Coleman, Miles Davis

We can make ‘$\exists x\,S(x)$’ true by including something in the extension of ‘$S$’, and we can make ‘$\forall x\,S(x)$’ false by leaving something out of the extension of ‘$S$’. For concreteness, let’s say:

$S(x)$:

blank${}_{x}$ plays saxophone

Now ‘$\exists x\,S(x)$’ is true, because ‘$S(x)$’ is true of Ornette Coleman. Slightly more precisely, extend our interpretation by allowing ‘$c$’ to name Ornette Coleman. ‘$S(c)$’ is true in this extended interpretation, so ‘$\exists x\,S(x)$’ was true in the original interpretation. Similarly, ‘$\forall x\,S(x)$’ is false, because ‘$S(x)$’ is false of Miles Davis. Slightly more precisely, extend our interpretation by allowing ‘$d$’ to name Miles Davis, and ‘$S(d)$’ is false in this extended interpretation, so ‘$\forall x\,S(x)$’ was false in the original interpretation. We have provided a counter-interpretation to the claim that ‘$\forall x\,S(x)$’ and ‘$\exists x\,S(x)$’ are logically equivalent.

To show that $\mathscr{A}$ and $\mathscr{B}$ are not logically equivalent, it suffices to find an interpretation where one is true and the other is false.

## 33.3 Validity, entailment and satisfiability

To test for validity, entailment, or satisfiability, we typically need to produce interpretations that determine the truth value of several sentences simultaneously.

Consider the following argument in FOL:

$\exists x(G(x)\rightarrow G(a))\therefore\exists x\,G(x)\rightarrow G(a)$

To show that this is invalid, we must make the premise true and the conclusion false. The conclusion is a conditional, so to make it false, the antecedent must be true and the consequent must be false. Clearly, our domain must contain two objects. Let’s try:

domain:

Karl Marx, Ludwig von Mises

$G(x)$:

blank${}_{x}$ hated communism

$a$:

Karl Marx

Given that Marx wrote The Communist Manifesto, ‘$G(a)$’ is plainly false in this interpretation. But von Mises famously hated communism, so ‘$\exists x\,G(x)$’ is true in this interpretation. Hence ‘$\exists x\,G(x)\rightarrow G(a)$’ is false, as required.

Does this interpretation make the premise true? Yes it does! Note that ‘$G(a)\rightarrow G(a)$’ is true. (Indeed, it is a validity.) But then certainly ‘$\exists x(G(x)\rightarrow G(a))$’ is true, so the premise is true, and the conclusion is false, in this interpretation. The argument is therefore invalid.

In passing, note that we have also shown that ‘$\exists x(G(x)\rightarrow G(a))$’ does not entail ‘$\exists x\,G(x)\rightarrow G(a)$’, i.e., that $\exists x(G(x)\rightarrow G(a))\nvDash\exists x\,G(x)\rightarrow G(a)$. Equally, we have shown that the sentences ‘$\exists x(G(x)\rightarrow G(a))$’ and ‘$\neg(\exists x\,G(x)\rightarrow G(a))$’ are jointly satisfiable.

Let’s consider a second example:

$\forall x\exists y\,L(x,y)\therefore\exists y\forall x\,L(x,y)$

Again, we want to show that this is invalid. To do this, we must make the premises true and the conclusion false. Here is a suggestion:

domain:

$L(x,y)$:

blank${}_{x}$ is in a domestic partnership with blank${}_{y}$

The premise is clearly true on this interpretation. Anyone in the domain is a Canadian citizen in a domestic partnership with some other Canadian citizen. That other citizen will also, then, be in the domain. So for everyone in the domain, there will be someone (else) in the domain with whom they are in a domestic partnership. Hence ‘$\forall x\exists y\,L(x,y)$’ is true. However, the conclusion is clearly false, for that would require that there is some single person who is in a domestic partnership with everyone in the domain, and there is no such person, so the argument is invalid. We observe immediately that the sentences ‘$\forall x\exists y\,L(x,y)$’ and ‘$\neg\exists y\forall x\,L(x,y)$’ are jointly satisfiable and that ‘$\forall x\exists y\,L(x,y)$’ does not entail ‘$\exists y\forall x\,L(x,y)$’.

For our third example, we’ll mix things up a bit. In chapter 30, we described how we can present some interpretations using diagrams. For example:

Using the conventions employed in chapter 30, the domain of this interpretation is the first three positive whole numbers, and ‘$R(x,y)$’ is true of x and y just in case there is an arrow from x to y in our diagram. Here are some sentences that the interpretation makes true:

• $\forall x\exists y\,R(y,x)$

• $\exists x\forall y\,R(x,y)$’ (witness for $x$: $1$)

• $\exists x\forall y(R(y,x)\leftrightarrow x=y)$’ (witness for $x$: $1$)

• $\exists x\exists y\exists z((\neg y=z\wedge R(x,y))\wedge R(z,x))$’ (witness for $x$: $2$)

• $\exists x\forall y\,\neg R(x,y)$’ (witness for $x$: $3$)

• $\exists x(\exists y\,R(y,x)\wedge\neg\exists y\,R(x,y))$’ (witness for $x$: $3$)

This immediately shows that all of the preceding six sentences are jointly satisfiable. We can use this observation to generate invalid arguments, e.g.:

 $\displaystyle\forall x\exists y\,R(y,x),\exists x\forall y\,R(x,y)$ $\displaystyle\therefore\forall x\exists y\,R(x,y)$ $\displaystyle\exists x\forall y\,R(x,y),\exists x\forall y\neg R(x,y)$ $\displaystyle\therefore\neg\exists x\exists y\exists z(\neg y=z\wedge(R(x,y)% \wedge R(z,x)))$

and many more besides.

If some interpretation makes all of $\mathscr{A}_{1},\mathscr{A}_{2},\ldots,\mathscr{A}_{n}$ true and $\mathscr{C}$ is false, then: 1.$\mathscr{A}_{1},\mathscr{A}_{2},\ldots,\mathscr{A}_{n}\therefore\mathscr{C}$ is invalid; and 2.$\mathscr{A}_{1},\mathscr{A}_{2},\ldots,\mathscr{A}_{n}\nvDash\mathscr{C}$; and 3.And $\mathscr{A}_{1},\mathscr{A}_{2},\ldots,\mathscr{A}_{n},\neg\mathscr{C}$ are jointly satisfiable.

An interpretation which refutes a claim—to logical truth, say, or to entailment—is called a counter-interpretation, or a counter-model.

We’ll close this section, though, with a caution about the relationship between (in)validity and (non)entailment. Recall FOL’s limitations: it is an extensional language; it ignores issues of vagueness; and it cannot handle cases of validity for ‘special reasons’. To take one illustration of these issues, consider this natural-language argument:

• Every fox is cute.

• All vixens are cute.

This is valid: necessarily every vixen is a fox, so it is impossible for the premise to be true and the conclusion false. Now, we might sensibly symbolize the argument as follows:

$\forall x(F(x)\rightarrow C(x))\therefore\forall x(V(x)\rightarrow C(x))$

However, it is easy to find counter-models which show that $\forall x(F(x)\rightarrow C(x))\nvDash\forall x(V(x)\rightarrow C(x))$. (Exercise: find one.) So, it would be wrong to infer that the English argument is invalid, just because there is a counter-model to the relevant FOL-entailment.

The general moral is this. If you want to infer from the absence of an entailment in FOL to the invalidity of some English argument, then you need to argue that nothing important is lost in the way you have symbolized the English argument.

## Practice exercises

A. Show that each of the following is neither a validity nor a contradiction:

1. 1.

$D(a)\wedge D(b)$

2. 2.

$\exists x\,T(x,h)$

3. 3.

$P(m)\wedge\neg\forall x\,P(x)$

4. 4.

$\forall zJ(z)\leftrightarrow\exists y\,J(y)$

5. 5.

$\forall x(W(x,m,n)\vee\exists y\,L(x,y))$

6. 6.

$\exists x(G(x)\rightarrow\forall y\,M(y))$

7. 7.

$\exists x(x=h\wedge x=i)$

B. Show that the following pairs of sentences are not logically equivalent.

1. 1.

$J(a)$, $K(a)$

2. 2.

$\exists x\,J(x)$, $J(m)$

3. 3.

$\forall x\,R(x,x)$, $\exists x\,R(x,x)$

4. 4.

$\exists x\,P(x)\rightarrow Q(c)$, $\exists x(P(x)\rightarrow Q(c))$

5. 5.

$\forall x(P(x)\rightarrow\neg Q(x))$, $\exists x(P(x)\wedge\neg Q(x))$

6. 6.

$\exists x(P(x)\wedge Q(x))$, $\exists x(P(x)\rightarrow Q(x))$

7. 7.

$\forall x(P(x)\rightarrow Q(x))$, $\forall x(P(x)\wedge Q(x))$

8. 8.

$\forall x\exists y\,R(x,y)$, $\exists x\forall y\,R(x,y)$

9. 9.

$\forall x\exists y\,R(x,y)$, $\forall x\exists y\,R(y,x)$

C. Show that the following sentences are jointly satisfiable:

1. 1.

$M(a),\neg N(a),P(a),\neg Q(a)$

2. 2.

$L(e,e),L(e,g),\neg L(g,e),\neg L(g,g)$

3. 3.

$\neg(M(a)\wedge\exists x\,A(x)),M(a)\vee F(a),\forall x(F(x)\rightarrow A(x))$

4. 4.

$M(a)\vee M(b),M(a)\rightarrow\forall x\neg M(x)$

5. 5.

$\forall y\,G(y),\forall x(G(x)\rightarrow H(x)),\exists y\neg I(y)$

6. 6.

$\exists x(B(x)\vee A(x)),\forall x\neg C(x),\forall x\bigl{[}(A(x)\wedge B(x))% \rightarrow C(x)\bigr{]}$

7. 7.

$\exists x\,X(x),\exists x\,Y(x),\forall x(X(x)\leftrightarrow\neg Y(x))$

8. 8.

$\forall x(P(x)\vee Q(x)),\exists x\neg(Q(x)\wedge P(x))$

9. 9.

$\exists z(N(z)\wedge O(z,z)),\forall x\forall y(O(x,y)\rightarrow O(y,x))$

10. 10.

$\neg\exists x\forall y\,R(x,y),\forall x\exists y\,R(x,y)$

11. 11.

$\neg R(a,a)$, $\forall x(x=a\vee R(x,a))$

12. 12.

$\forall x\forall y\forall z[(x=y\vee y=z)\vee x=z]$, $\exists x\exists y\ \neg x=y$

13. 13.

$\exists x\exists y((Z(x)\wedge Z(y))\wedge x=y)$, $\neg Z(d)$, $d=e$

D. Show that the following arguments are invalid:

1. 1.

$\forall x(A(x)\rightarrow B(x))\therefore\exists x\,B(x)$

2. 2.

$\forall x(R(x)\rightarrow D(x)),\forall x(R(x)\rightarrow F(x))\therefore% \exists x(D(x)\wedge F(x))$

3. 3.

$\exists x(P(x)\rightarrow Q(x))\therefore\exists x\,P(x)$

4. 4.

$N(a)\wedge N(b)\wedge N(c)\therefore\forall x\,N(x)$

5. 5.

$R(d,e),\exists x\,R(x,d)\therefore R(e,d)$

6. 6.

$\exists x(E(x)\wedge F(x)),\exists x\,F(x)\rightarrow\exists x\,G(x)\therefore% \exists x(E(x)\wedge G(x))$

7. 7.

$\forall x\,O(x,c),\forall x\,O(c,x)\therefore\forall x\,O(x,x)$

8. 8.

$\exists x(J(x)\wedge K(x)),\exists x\neg K(x),\exists x\neg J(x)\therefore% \exists x(\neg J(x)\wedge\neg K(x))$

9. 9.

$L(a,b)\rightarrow\forall x\,L(x,b),\exists x\,L(x,b)\therefore L(b,b)$

10. 10.

$\forall x(D(x)\rightarrow\exists y\,T(y,x))\therefore\exists y\exists z\ \neg y=z$