# First order logic - is this a valid interpretation of a sentence?

In the book I'm reading, the following argument is presented

There are liars. Therefore, there are thieves.

The premise is symbolised as ∃x(Lx) while the conclusion is symbolised as ∃x(Tx). The following interpretation is given to show that the argument is invalid:

Let the domain be {1,3} and the predicates L(x) and T(x) be interpreted as :

L(x) ↔ x = 1

T(x) ↔ x = 2

So that the premise ∃x(Lx) is true when x = 1 whereas the conclusion ∃x(Tx) is false with respect to the domain.

My confusion comes from '2' being in the predicate T(x). 2 isn't in the domain and I'd like to know if interpretations such as this is really allowed in first order logic. Sure the interpretation 'x=2' has the same number of free variables as T(x), but since 2 isn't in the domain, shouldn't 'x=2' technically have no meaning in this domain and therefore no truth value?

• Obviously, we can interpret variables with elements in the domain; but also constants must refer to elements in the domain. Thus, what is the meaning of the constant $2$ ?. – Mauro ALLEGRANZA Mar 13 at 7:16
• Consider as domain the set $\mathbb N$ of naturals with $0$. Let $L(x)$ interpreted with "$x$ is greater-or-equal to $0$" and let $T(x)$ interpreted with "$x$ is less than $0$". – Mauro ALLEGRANZA Mar 13 at 7:18
• So you mean to say that the interpretation 'x=2' is not a valid interpretation when the domain is {1,3} (which excludes 2)? This is what I thought essentially. I was just surprised this appeared in the author's example and I started questioning myself (the author is quite knowledgeable you see) – Unknowledgeable Mar 13 at 7:28
• In my interpretation, it's either a typo ($3$ could be $2$), or '$x=2$' is simply considered a valid statement which never holds on the given domain. – Berci Mar 13 at 7:35
• Ah I see what you mean... I can make an interpretation like 'x = elephant' and it would be a valid interpretation except that it would be false. I think I got it. Thanks! – Unknowledgeable Mar 13 at 7:44

You are right. Normally, in a structure with domain $$\mathcal{A}$$ and $$\mathcal{I}$$, for an $$n$$-ary predicate $$P$$, $$\mathcal{I}(P^n) \subseteq \mathcal{A}^n$$, i.e. interpretation of an 1-ary predicate is a subset of the domain, the interprtation of an $$2$$-ary predicate is a subset of $$\mathcal{A} \times \mathcal{A}$$, ...,
Your book might presuppose some weird definition of models in which non-logical symbols can be interpreted as anything, but normally you'd want models to be closed systems, in the sense that interpretating some predicate shouldn't shoot you out of the domain of the model.

In order to prove that $$\exists x L(x) \not \vDash \exists x T(x)$$, they should rather have presented a model in which there are liears but no thieves at all, e.g.

$$\text{domain} = \{1\};\ \mathcal{I}(L) = \{1\};\ \mathcal{I}(T) = \emptyset$$

Or, with the $$\leftrightarrow$$ notation, $$L(x) \leftrightarrow x = 1;\ T(x) \leftrightarrow \bot$$ - but this notation presupposes that 1 (and 2 etc.) are constants, which, as Mauro Allegranza points out in their comment, need to be assigned an interpretation as well in order to be any meaningful.

May I ask which book you are using, and how they define models? If they nowhere give a precise definition of what an interpretation of a predicate is, then it's probably not a good book anyway.

• I actually have not yet learned models (and the book hasn't introduced them) so I don't fully understand your explanation yet. I do agree that they should have presented a model with liars but no thieves though. Thanks for your answer, I'll defo come back to it after learning about models. – Unknowledgeable Mar 13 at 8:01
• The book I'm reading is old actually. Introduction to Logic by Patrick Suppes. I think it gave some new contributions to Logic at the time of its publication – Unknowledgeable Mar 13 at 8:02
• Regarding the definition of an interpretation, it offers the following: "Sentence P is an interpretation of formula Q with respect to the domain of individuals D if and only if P can be obtained from Q by substituting predicates and operation symbols defined for the individuals in the domain D for the predicates and operation symbols respectively of Q and by substituting proper names of individuals in D for proper names (i.e., individual constants) and free variables of Q." Since 'x=2' is a predicate not defined in the domain, I guess it's an invalid interpretation as suspected – Unknowledgeable Mar 13 at 8:37
• Ignore my previous, now deleted comment. The crucial part is "substituting [...] proper names of individuals in D for [...] free variables of Q." According to this, 2 needs to be a proper name. So whether "x=2" is a proper name depends on whether "2" is a proper name in the language that has itself an interpretaton, which should again be in the domain D. Its semantic validity in turn depends on whether that interpretation makes the argument valid; but with the interpreation given, the only way to refute $\exists x T(x)$ is to not have any elements in the interpration of T at all, ... – lemontree Mar 13 at 8:49
• ... so no matter what individual the proper name "2" is supposed to refer to, given that it has to be something from the domain (1 or 3), either interpretation will not support the conclusion. So my judgement is that this is a syntactically valid definition, provided that "2" is a proper name, but semantically nonsensical, provided that the interpreation of "2" has to be in the domain $\{1,3\}$. – lemontree Mar 13 at 8:49

Your counter-example with domain $$D = \{ 1,3 \}$$ can be misleading, due to the fact that the number two is not in $$D$$...

It works when we interpret $$L(x)$$ as the property "x is equal to one" and $$T(x)$$ as the property "x is equal to two".

We have to carefully avoid the conflation of "objects" and "names": according to the semantics for first-order logic we can have objects (i.e. elements of the domain) without name (i.e. without individual constants referring to them) but we cannot have constants without reference.

Another counter-example can be based on the following interpretation for $$L(x)$$ and $$T(x)$$ respectively :

"$$x \text { is Odd}$$" and "$$x \text { is Even}$$".

In this case, the argument will become:

There are Odd numbers. Therefore, there are Even numbers.

which is clearly falsified in $$\{ 1,3 \}$$.

But, IMO, the examples above are not based on what the author call: arithmetical interpretation, i.e. an interpretation in the domain of positive integers (see Patrick Suppes, Introduction to Logic, page 64).

An arithmetical counter-example, based on domain $$\mathbb N = \{ 1, 2, \ldots \}$$, will be the following:

let $$L(x)$$ interpreted as $$(x \ge 1)$$ and let $$T(x)$$ interpreted as $$(x < 1)$$.

Having found suitable interpretations showing that: $$∃xL(x) \nvDash ∃xT(x)$$, we have showed that the argument:

There are liars. Therefore, there are thieves.

is not valid.

• Really appreciate your answers, although my main trouble was deciding whether or not 'x=2' is a valid interpretation because 2 is outside the domain – Unknowledgeable Mar 13 at 8:13
• Ah I think I get it. The interpretation is valid for as long as I assign 2 to mean something in the domain (either 1 or 3). Thanks! – Unknowledgeable Mar 13 at 8:47

When it says:

Let the domain be {1,3} and the predicates L(x) and T(x) be interpreted as :

L(x) ↔ x = 1

T(x) ↔ x = 2

it is not considering $$x=2$$ to be a logic formula that is yet to be interpreted, but rather it indicates for which objects of the domain the predicate holds true (so yes, that's very confusing; they mix up logic notation with mathematical expressions about the interpretation)

So, it says that the predicate $$T$$ holds for object $$2$$ ... which is not part of the domain .. and so in effect there are no thieves at all. Which is what you want, since object $$1$$ is a liar, and therefore the premise is true, and the conclusion is false, and thus we have a counterexample, as desired.

What I don;t understand, though, is that they could simply have picked the domain as $$\{ 1 \}$$ ... that would have worked just as well.