Let us have some formal language $\mathcal{L}$ and an $\mathcal{L}$-structure $\mathcal{U}=(A,\mathcal{I})$. Where $A$ - non-empty set, called domain, and $\mathcal{I}$ - interpretation.

I know that interpretation is a function (or family of functions) that assigns to every constant, functional or predicative symbol from the signature of the language $\mathcal{L}$ correspondent elements, functions and relations on the domain $A$.

Also, I know that there exists some function $\mu : T \to A$ called variable assignment that associates every (variable) term of the language $\mathcal{L}$ to the element of domain $A$.

Question 1: Where does this function $\mu$ belong? Is it the part of the structure $\mathcal{U}$ or interpretation $\mathcal{I}$ ? Or is it part of a formal language? As it is dependent on a domain, why it is not included in the definition of a structure $\mathcal{U}$?

Then, if we want to assign "truth-values" to every sentence of our language $\mathcal{L}$, we need to specify a so-called object of truth values $\Omega$ and a truth-valuation function $V:\Phi\to \Omega$ (where $\Phi$ is a set of all sentences of $\mathcal{L}$).

Question 2: Similar to the first. Does this $\Omega$ or $V$ has anything to do with the domain $A$ or interpretation $\mathcal{I}$ of the structure $\mathcal{U}$, or with variable assignment function $\mu$? Or they are purely independently defined objects? I thought that assignment of truth-values should be dependent on the structure in which we want to interpret our language, (for example, I thought that $\Omega$ might be an element or a subset of $A$) but I do not clearly see the connection between those notions.


An interpretation $\mathcal I$ is a function from the set of symbols of the language to "objects of the world", i.e. to elements of the domain $A$ or of its power-set.

In the same way, a variable assignment function $\mu$ is a function from a subset of symbols of the language, i.e. the set of variables to the domain $A$, i.e.:

$\mu : \text{Var} \to A$.

The variable assignment function "trick" is used to handle open formulas, like e.g. $(x=0)$.

How can we determine the truth-value of $(x=0)$ in the structure $\mathbb N$ ? It depends on what $x$ refers to.

A free var in a formula acts as a pronoun; when I assert "it is red", the truth value of my assertion will depend on what "it" refers to: if it refers to my car, the statement asserted is false, if it refers to my pen, it is true.

In the same way, the interpretation will define the denotation of the constant $0$: the number zero.

With a variable assignment function $\mu$ such that $\mu(x)=0$ we have that the formula $(x=0)$ is satisfied in $\mathbb N$ with the variable assigment fucntion $\mu$, i.e.

$\mathbb N, \mu \vDash (x=0)$.

With e.g. a different function $\mu'$ such that $\mu'(x)=1$, obviously the above formula is not satisfied.

More formally, for an atomic formula $x=c$, with $c$ an individual constant, we have that it is satisfied in $\mathcal A$ with $\mu$ (i.e. $\mathcal A, \mu \vDash (x=c)$) iff the element of $A$ (domain of $\mathcal A$) assigned to $c$ as referent, call it $c^A$, is the same element assigned to $x$ by $\mu$, i.e. $μ(x)=c^A$.

In conclusion, the interpretation and a variable assignment function determine univocally a truth value for every formula of the language. If the formula is a closed one (i.e. a sentence) its truth value does not depends on the variable assignment.

  • $\begingroup$ Thanks! How do we formally define if our formula "is satisfied"? I think we must somehow formally "assign" some truth-value to it. But how specifically is this assignation done? Also, what if I want to have not only "TRUE" and "FALSE" values, but few others too? $\endgroup$ – Sergey Dylda Jun 29 '17 at 10:59
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    $\begingroup$ @SergeyDylda - the formula $\varphi(x)$ is satisfied in the interpretation $\mathcal A$ with var assignment $\mu$ (in symbols: $\mathcal A, \mu \vDash \varphi(x)$) iff ... several clauses. Example: if $\varphi(x)$ is the atomic formula $x=c$ with $c$ individual constant, it is satisfied iff the element of $A$ (domain of $\mathcal A$) assigned to $c$ as referent, call it $c^A$, is the same element assigned to $x$ by $\mu$, i.e. $\mu(x)=c^A$. $\endgroup$ – Mauro ALLEGRANZA Jun 29 '17 at 11:54

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