An Update of Tarski: Two Usages of the Word “True”

Abstract

This paper is based on Tarski’s theory of truth. The purpose of this paper is to solve the liar paradox (and its cousins) and keep both of the deductive power of classical logic and the expressive power of the word “true” in natural language. The key of this paper lies in the distinction between the predicate usage and the operator usage of the word “true”. The truth operator is primarily used for characterizing the semantics of the language. Then, we do not need the hierarchy of languages. The truth predicate is mainly used for grammatical function. Tarski’s schema of the truth predicate is not necessary in this proposal. The schema of the word "true" is the schema of the truth operator. The liar paradox (and its cousins) can be solved in this way. In the appendix, I show a consistent model for both of the truth predicate and the truth operator.

Appendix: A Model of the Two Usages of the Word “True”

How to define exactly the extensions of the two usages of the word “true”? It is better to talk about them in a formalized language. In this appendix, I use the methods of the revision theory to construct a model to characterize precisely the main ideas of this paper.²⁴

In order to express sentences in natural language, let the base language $\mathcal {L}$ be Peano Arithmetic (PA) plus some other terms and predicates which are different from the word “true”. Let $\mathcal {L_{T}}$ be the language obtained by adding a one place predicate $T_{p}$ and a unary operator $T_{o}$ to $\mathcal {L}$, in which $T_{p}$ is the truth predicate and $T_{o}$ is the truth operator.

1.1 The Syntax of $\mathcal {L_{T}}$

The construction of the well-formed formulas of $\mathcal {L}$ is as usual. The construction of the well-formed formulas of the language $\mathcal {L_{T}}$ is based on the construction of $\mathcal {L}$:

• If t is a term of $\mathcal {L}$, so it is also a term of $\mathcal {L_{T}}$, then $T _{p}$(t) is a well-formed formula

• If $\phi$ is a formula of $\mathcal {L_{T}}$, then $T _{o}$($\phi$) is a formula of $\mathcal {L_{T}}$.

Other well-formed formulas of $\mathcal {L_{T}}$ are defined as usual, i.e.:

• If $t_{1}$, $t_{2}$, $t_{3} \ldots t_{n}$ are terms of $\mathcal {L_{T}}$ and $P_{n}$ is a n-ary predicate of $\mathcal {L_{T}}$, which is different from T$_{p}$, then $P_{n}$( $t_{1}$, $t_{2}$, $t_{3} \ldots t_{n}$) is a well-formed formula of $\mathcal {L_{T}}$.

• If $\phi$ is a formula of $\mathcal {L_{T}}$, then $\lnot \phi$, $\forall x\phi$ and $\exists x\phi$, are well-formed formulas of $\mathcal {L_{T}}$.

• If $\phi$ and $\psi$ are formulas of $\mathcal {L_{T}}$, then $\phi \vee \psi$, $\phi \wedge \psi$, $\phi \rightarrow \psi$, $\phi \leftrightarrow \psi$ are well-formed formulas of $\mathcal {L_{T}}$.

1.2 The Semantics of $\mathcal {L_{T}}$

$\mathcal {M}_0 = \langle \Delta _0, \sigma _0 \rangle$ is the initial model of $\mathcal {L}_{T}$. $\Delta _0$ is the domain, and $\sigma _0$ is the assignment function which contains the standard interpretation of PA.

• For a term t: $\sigma _0(t) \in \Delta _0$

• For an n-ary predicate P$_{n}$: $\sigma _0(P_{n}) \subseteq \Delta _{0}^n$

• The evaluations of the operators and formulas of $\mathcal {L}_{T}$ are as usual.

$\mathcal {M}_{\alpha +1} = \langle \Delta _{\alpha +1}, \sigma _{\alpha +1} \rangle$, where $\Delta _{\alpha +1} =\Delta _0$, is the expansion of $\mathcal {M}_\alpha$ by adding the interpretations of $T_{p}$ and $T _{o}$ as follows:

• $\sigma _{\alpha +1}$($T _{p}$)=$\{ \ulcorner \phi \urcorner : \sigma _{\alpha }\models \phi \}$, where $\phi$ is a well-formed sentence of $\mathcal {L}_{T}$ and $\ulcorner \phi \urcorner$ is the Gödel number of $\phi$.

• $\sigma _{\alpha +1}$($T _{o} \phi$)=$\sigma _{\alpha +1}$($\phi$), where $\phi$ is a sentence of $\mathcal {L}_{T}$.

For limit ordinal $\lambda$, $\mathcal {M}_{ \lambda } = \langle \Delta _{ \lambda }, \sigma _{ \lambda } \rangle$, where $\Delta _{\lambda } =\Delta _0$, is defined as follows:

• For a term t: $\sigma _\lambda (t)= \sigma _0(t) \in \Delta _0$

• For an n-ary predicate P$_{n}$, which is different from T$_{p}$, $\sigma _\lambda (P_{n})=\sigma _0(P_{n}) \subseteq \Delta _{0}^n$

• $\sigma _\lambda (T_p) = \{\ulcorner \phi \urcorner : \text { there exists an } \alpha< \lambda \text { such that } \mathcal {M}_\alpha \vDash \phi \text { and, for any } \alpha< \beta < \lambda , \mathcal {M}_\beta \vDash \phi \}$

• $\sigma _{\lambda }$($T _{o} \phi$)=$\sigma _{\lambda }$($\phi$),

• The evaluations of other operators and formulas are as usual.

$\mathcal {M} = \langle \Delta , \sigma \rangle$ is the final model of $\mathcal {L}_{T}$ where $\Delta$=$\Delta _{0}$. It is defined as follows:

• For a term t: $\sigma (t)= \sigma _0(t) \in \Delta _0$

• For an n-ary predicate P$_{n}$, which is different from $T _{p} , \sigma (P_{n})=\sigma _0(P_{n}) \subseteq \Delta _{0}^n$

• $\sigma (T_p) = \{\ulcorner \phi \urcorner :$ there exists a limit ordinal $\lambda$ such that $\mathcal {M}_\lambda \vDash \phi$ and, for any limit ordinal $\nu > \lambda , \mathcal {M}_\nu \vDash \phi \}$

• $\sigma$($T_{o} \phi$)=$\sigma$($\phi$)

• The evaluations of other operators and formulas are as usual.

Case 1. First, let us talk about the liar sentence $\lnot T_{p} (\ulcorner \phi \urcorner )$, where $\phi$ is the abbreviation of the sentence $\lnot T_{p} (\ulcorner \phi \urcorner )$ itself and $\ulcorner \phi \urcorner$ is the name of $\phi$.

Suppose we talk about this example in the final model. Because there is no $T_{p}$ in $\mathcal {L}$, $\phi$ is not a sentence of $\mathcal {L}$. Hence it is not true that $\sigma _{0} \models \phi$. Then it is not true that $\ulcorner \phi \urcorner \in \sigma _{1}(T_{p})$, so $\sigma _{1}(T_{p} ( \ulcorner \phi \urcorner ))=0$. Hence $\sigma _{1}(\lnot T_{p} ( \ulcorner \phi \urcorner ))=1$. Therefore $\ulcorner \phi \urcorner \in \sigma _{2}(T_{p})$ because $\phi =\lnot T_{p} ( \ulcorner \phi \urcorner )$. Then $\sigma _{2}(T_{p}( \ulcorner \phi \urcorner ))=1$. Hence $\sigma _{2}(\lnot T_{p}( \ulcorner \phi \urcorner ))=0$.

With the iteration of the models, we can conclude that $\lnot T_{p} ( \ulcorner \phi \urcorner )$ is true in the models $\sigma _{2n+1}$ and false in the models $\sigma _{2n+2}$, where $n\in \omega$. So there is no n, such that for any $k>n$, $\sigma _{k}$($\lnot T_{p} ( \ulcorner \phi \urcorner )$)=1. Hence, $\ulcorner \phi \urcorner \notin \sigma _{\omega }(T_{p})$. And then $\sigma _{\omega }(T_{p} ( \ulcorner \phi \urcorner ))=0$, therefore $\sigma _{\omega }(\lnot T_{p} ( \ulcorner \phi \urcorner ))=1$.

The iteration of the values of the liar sentence can be seen in the following Fig. 1. It is not hard to see that the liar sentence $\lnot T_{p} (\ulcorner \phi \urcorner )$ is true in every limit model. So in the final model $\mathcal {M} = \langle \Delta , \sigma \rangle$, $\ulcorner \phi \urcorner \in \sigma (T_{p})$. Then $\sigma (T_{p} ( \ulcorner \phi \urcorner ))=1$, therefore $\sigma (\lnot T_{p} ( \ulcorner \phi \urcorner ))=0$.

Fig. 1

The revision sequence of the liar sentence

Case 2. For the truth teller sentence $T_{p} (\ulcorner \psi \urcorner )$, where $\psi$ is the abbreviation of the sentence $T_{p} (\ulcorner \psi \urcorner )$ itself and $\ulcorner \psi \urcorner$ is the name of $\psi$. Since $\psi$ is not a sentence of $\mathcal {L}$, it is not true that $\sigma _{0}\models \psi$. Hence $\ulcorner \psi \urcorner \notin \sigma _{1}(T_{p})$, so $\sigma _{1}(T_{p}( \ulcorner \psi \urcorner ))=0$. Then $\ulcorner \psi \urcorner \notin \sigma _{2}(T_{p})$. So $\sigma _{2}(T_{p}( \ulcorner \psi \urcorner ))=0$. It is obvious that with the iteration of the models, the truth teller sentence $T_{p} (\ulcorner \psi \urcorner )$ is always false.

The evaluations of the truth teller sentence can be seen in the following Fig. 2. So in the final model $\mathcal {M} = \langle \Delta , \sigma \rangle$, $\ulcorner \psi \urcorner \notin \sigma (T_{p})$. Then $\sigma (T_{p} ( \ulcorner \psi \urcorner ))=0$. Therefore, $T_{p} ( \ulcorner \psi \urcorner )\leftrightarrow \psi$.

Fig. 2

The revision sequence of the truth teller sentence

Definition 1

A sentence $\phi$ of $\mathcal {L_{T}}$ is $\mathcal {L_{T}}$-valid iff for any final model $\tau$ of $\mathcal {L_{T}}$, $\tau \models _{ \mathcal {L_{T}} } \phi$.²⁵

Then we can prove that the deductive power of classical logic is kept in $\mathcal {L_{T}}$:

Theorem 1

For any sentence $\phi$ of classical logic, if $\phi$ is valid in classical logic, $\phi$ is $\mathcal {L_{T}}$-valid.

Proof

Suppose there is a sentence $\psi$ that is valid in classical logic but not $\mathcal {L_{T}}$-valid. Then there is a $\mathcal {L_{T}}$ model $\tau$ such that $\tau (\psi ) =0$. Since, according to the construction of the models for $\mathcal {L_{T}}$, the evaluations of the normal operators in any $\mathcal {L_{T}}$ model are as usual, we can construct a model $\sigma$ of classical logic which has the same evaluations as that of $\tau$ for the operators appear in $\psi$. Then $\psi$ is false in $\sigma$, i.e. $\sigma (\psi ) =0$. Contradiction. $\square$

Then we can prove that Tarski’s T-schema for the truth predicate is not a $\mathcal {L_{T}}$-valid schema:

Theorem 2

$T_{p} (\ulcorner \phi \urcorner )\leftrightarrow \phi$ is not $\mathcal {L_{T}}$-valid.

Proof

As we can see in the above Case 1, there exists at least one sentence $\phi$, e.g. the liar sentence $\lnot T_{p} (\ulcorner \phi \urcorner )$, such that in the final model $\mathcal {M} = \langle \Delta , \sigma \rangle$, $\sigma ( \phi )$= $\sigma (\lnot T_{p} (\ulcorner \phi \urcorner )) =0$, but $\sigma (T_{p} (\ulcorner \phi \urcorner )) =1$. Hence, $T_{p} (\ulcorner \phi \urcorner )\leftrightarrow \phi$ is not $\mathcal {L_{T}}$-valid. $\square$

But we can prove that the schema for the truth operator, i.e. T$_{o}$-schema, is $\mathcal {L_{T}}$-valid:

Theorem 3

$T_{o}(\phi ) \leftrightarrow \phi$ is $\mathcal {L_{T}}$-valid, i.e. for any final model $\tau$, $\tau \models _{ \mathcal {L_{T}} } T_{o}(\phi ) \leftrightarrow \phi$.

Proof

Directly from the definition of $T_{o}$. $\square$

Corollary 1

For any sentence $\phi$, if we can equivalently translate $T_{p} (\ulcorner \phi \urcorner )$ to a well-formed sentence $T_{o}( \phi )$ , then $T_{p} (\ulcorner \phi \urcorner )\leftrightarrow \phi$ is $\mathcal {L_{T}}$-valid and vice versa i.e. the following T$_{op}$-schema is $\mathcal {L_{T}}$-valid:

(T$_{op}$) $(T_{p} (\ulcorner \phi \urcorner )\leftrightarrow T_{o}(\phi ) ) \leftrightarrow (T_{p} (\ulcorner \phi \urcorner )\leftrightarrow \phi )$

Proof

For any final model $\tau$ and for any sentence $\phi$, suppose $T_{p} (\ulcorner \phi \urcorner )\leftrightarrow T_{o}(\phi )$ is true in $\tau$. According to Theorem 3, $T_{o}(\phi ) \leftrightarrow \phi$ is true in $\tau$. Hence, $\tau (T_{p} (\ulcorner \phi \urcorner ))$=$\tau (T_{o}(\phi ))$ and $\tau (T_{o}(\phi ))$=$\tau (\phi )$. Then $\tau (T_{p} (\ulcorner \phi \urcorner ))$=$\tau (\phi )$. Therefore, $T_{p} (\ulcorner \phi \urcorner )\leftrightarrow \phi$ is true in $\tau$. The other direction is similar. $\square$

From Theorems 2 and 3, we can conclude that not every instance of an operator usage of the word “true” is equivalent to the corresponding instance of the predicate usage. For example, suppose (L) is the liar sentence, i.e. “($\ulcorner L \urcorner$) is not true”, then the value of the sentence “it is true that ($\ulcorner L \urcorner$) is not true”²⁶ is equivalent to the value of the sentence (L), namely it is false. But the value of the sentence “‘($\ulcorner L \urcorner$) is not true’ is true” is true.²⁷ Therefore, not every translation from an operator usage into a predicate usage of the word “true” is equivalent, although every operator usage of “true” can be translated into a predicate usage uniformly. This is the reason why the operator usage of “true” is not equivalent to (part of) the predicate usage of “true”. On the other hand, not every truth predicate can be translated uniformly into an operator usage of the word “true”, let alone an equivalent translation. Hence the truth operator and the truth predicate are two different usages of the word “true”. They have different functions, although they have some connections. In short, the truth operator is mainly used for characterizing the semantics of the language and the truth predicate is mainly used for generalizations. We cannot reduce one usage to the other.

Definition 2

For any sentence $\phi$ and $\psi$ of $\mathcal {L_{T}}$, $\phi$ and $\psi$ are $\mathcal {L_{T}}$-equivalent iff for any final model $\tau$ of $\mathcal {L_{T}}$, $\tau \models _{ \mathcal {L_{T}} } \phi \leftrightarrow \psi$.

Corollary 2

For any sentence $\phi$, the T$_{o}$-schema and the T$_{op}$-schema are $\mathcal {L_{T}}$-equivalent, i.e. for any final model $\tau$ of $\mathcal {L_{T}}$:

$\tau \models _{ \mathcal {L_{T}} }(T_{o}(\phi ) \leftrightarrow \phi )\leftrightarrow ((T_{p} (\ulcorner \phi \urcorner )\leftrightarrow T_{o}(\phi ) ) \leftrightarrow (T_{p} (\ulcorner \phi \urcorner )\leftrightarrow \phi ))$

Proof

Straightforward. $\square$

Definition 3

A unary logical operator $\odot$ weakly characterize a semantic status $\xi$ iff for any sentence $\phi$:

• $\bigodot (\phi )$ receives 1 iff $\phi$ receives $\xi$

Definition 4

A unary logical operator $\odot$ strongly characterize a semantic status $\xi$ iff for any sentence $\phi$:

• $\bigodot (\phi )$ receives 1 if $\phi$ receives $\xi$, and

• $\bigodot (\phi )$ receives 0 otherwise.

It is not hard to see that the weak characterization is equivalent to the strong characterization in bivalent logic. Hence, for $\mathcal {L}_{T}$, the weak characterization and the strong characterization are equivalent. Then we can prove that:

Theorem 4

The operator $T_{o}$ weakly (and strongly) characterize the semantic statuses of the language $\mathcal {L}_{T}$.

Proof

For any sentence $\phi$ of $\mathcal {L}_{T}$, for any final model $\tau$ of $\mathcal {L}_{T}$, according to the definition of the operator $T_{o}$, it is obvious that:

$$\begin{aligned} \tau (T_{o} (\phi ))=1\hbox { iff } \tau (\phi )=1 \end{aligned}$$

Case 3. It is very interesting to talk about the truth value of the sentences in example (2). Here we formalize example (2) as follows:

$$\begin{aligned} (2\hbox {F})&\phi : T_{p} (\ulcorner \psi \urcorner ) \\&\psi : \lnot T_{p} (\ulcorner \phi \urcorner ) \end{aligned}$$

Consider this example in the final model. Since there is no $T_{p}$ in $\mathcal {L}$, $\phi$ and $\psi$ are not sentences of $\mathcal {L}$. Then it is not true that $\sigma _{0}\models \phi$ and, it is not true that $\sigma _{0}\models \psi$. Hence it is not true that $\ulcorner \phi \urcorner \in \sigma _{1}(T_{p})$, and it is not true that $\ulcorner \psi \urcorner \in \sigma _{1}(T_{p})$. Then $\sigma _{1}(T_{p}(\ulcorner \phi \urcorner )) =0$ and $\sigma _{1}(T_{p}(\ulcorner \psi \urcorner )) =0$. Therefore $\sigma _{1}(\lnot T_{p}(\ulcorner \phi \urcorner )) =1$, i.e. $\sigma _{1}(\phi ) =0$ and $\sigma _{1}(\psi ) =1$. Then it is not true that $\ulcorner \phi \urcorner \in \sigma _{2}(T_{p})$ while it is true that $\ulcorner \psi \urcorner \in \sigma _{2}(T_{p})$. Then $\sigma _{2}(T_{p}(\ulcorner \psi \urcorner )) =1$ and $\sigma _{2}(T_{p}(\ulcorner \phi \urcorner )) =0$. Hence, $\sigma _{2}(\lnot T_{p}(\ulcorner \phi \urcorner )) =1$, i.e. $\sigma _{2}(\phi ) =1$ and $\sigma _{2}(\psi ) =1$. Then in $\sigma _{3}$, $\sigma _{3}(\phi ) =1$ and $\sigma _{3}(\psi ) =0$. And in $\sigma _{4}$, $\sigma _{4}(\phi ) =0$ and $\sigma _{4}(\psi ) =0$. In $\sigma _{5}$, $\sigma _{5}(\phi ) =0$ and $\sigma _{5}(\psi ) =1$ and so on. The evaluations of the two sentences in example (2) are as follows:

Fig. 3

The revision sequences of the sentences in (2F) in finite stages

As we can see in Fig. 3, there does not exist an n such that for any $k>n$, $\sigma _{k} (\phi )=1$, and there does not exist a m such that for any $s>m$, $\sigma _{s} (\psi )=1$. Then $\ulcorner \phi \urcorner \notin \sigma _{\omega } (T_{p})$ and $\ulcorner \psi \urcorner \notin \sigma _{\omega } (T_{p})$. Hence $\sigma _{\omega } (\phi ) = \sigma _{\omega } (T_{p}(\ulcorner \psi \urcorner ))=0$, while $\sigma _{\omega } (\psi ) = \sigma _{\omega } (\lnot T_{p}(\ulcorner \phi \urcorner ))=1$. The evaluations of the two sentences will go on as we can see in the following Fig. 4. And it is not hard to find out that the sentence $\phi$ is false in every limit model and the sentence $\psi$ is true in every limit model. Hence, in the final model, $\sigma (\phi ) = \sigma (T_{p} (\ulcorner \psi \urcorner ))=1$ and $\sigma (\psi )=\sigma ( \lnot T_{p} (\ulcorner \phi \urcorner )) =1$.

Fig. 4

The revision sequences of the sentences in (2F) in infinite stages

Fig. 5

The revision sequences of the Yablo sentences

Case 4. Yablo’s paradox can also be handled in this proposal. Consider Yablo’s paradox in the final model. Because there is no sentence containing the predicate $T_{p}$ or the operator $T_{o}$ in $\mathcal {L}$, for any sentence $S_{n}$ in the sequence of Yablo’s paradox, which can be formalized as $\forall x((x>n) \rightarrow \lnot T_{p}(\ulcorner S_{x} \urcorner ))$, $n\in \omega$, it is not true in the model $\sigma _{0}$. Hence for any $n\in \omega$, it is not true that $\ulcorner S_{n} \urcorner \in \sigma _{0}(T_{p})$. Therefore, for any $n\in \omega$, $\sigma _{1}(\lnot T_{p}(\ulcorner S_{n} \urcorner )) =1$. Hence for any $n \in \omega$, $\sigma _{1}(S_{n}) =1$. So for any $n \in \omega$, $\ulcorner S_{n} \urcorner \in \sigma _{2}(T_{p})$. Then for any $n\in \omega$, $\sigma _{2}(T_{p}(\ulcorner S_{n} \urcorner )) =1$. Therefore, for any $n\in \omega$, $\sigma _{2}(\lnot T_{p}(\ulcorner S_{n} \urcorner )) =0$. Then it is not true that for any $x>n$, $\sigma _{2}(\lnot T_{p}(\ulcorner S_{x} \urcorner )) =1$. Hence, for any $n\in \omega$, $\sigma _{2}(S_{n})=0$. Just like the liar sentence, the evaluations of the sentences in Yablo’s paradox alternate with the iteration of models. And hence for any $n \in \omega$, $\ulcorner S_{n} \urcorner \notin \sigma _{\omega }(T_{p})$. So for any $n \in \omega$, $\sigma _{\omega }(\lnot T_{p}(\ulcorner S_{n} \urcorner )) =1$ i.e. for any $n \in \omega$, $\sigma _{\omega }(S_{n} ) =1$. It is easy to see that for any $n \in \omega$, for any limit ordinal $\lambda$, $\sigma _{\lambda }(S_{n} ) =1$. The iterations of the sentences in Yablo’s paradox can be seen in Fig. 5. Therefore, in the final model, for any $n \in \omega$, $\ulcorner S_{n} \urcorner \in \sigma (T_{p})$. And so, for any $n \in \omega$, $\sigma ( T_{p}(\ulcorner S_{n} \urcorner )) =1$. Hence, for any $n \in \omega$, $\sigma (\lnot T_{p}(\ulcorner S_{n} \urcorner )) =0$ i.e. for any $n \in \omega$, $\sigma (S_{n} ) =0$.