# Example of complete formula on constants.

I have question considering example of complete formula, that was presented in Keisler/Chang - Model Theory:

2.3.1. EXAMPLES

(1). Let $$T$$ be a complete theory and let $$c_0,c_1,c_2,\ldots$$ be constant symbols of $$\mathscr{L}$$. Then any formula of $$\mathscr{L}$$ of the form $$x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$$ is complete in $$T$$. If $$\mathfrak{U}$$ is a model of $$T$$ such that every element of $$A$$ is a constant, then $$\mathfrak{U}$$ is an atomic model.

Why so? Am I right, when I think, that if theory is complete than -> every of it's formulas is complete too? So that formula with conjunction too?

Also, I pin basic definitions about it:

Consider a complete theory $$T$$ in $$\mathcal{L}$$. A formula $$\varphi(x_1\ldots x_n)$$ is said to be complete (in $$T$$) iff for every formula $$\psi(x_1,\ldots x_n)$$ exactly one of $$T\vDash\varphi\rightarrow\psi,\quad T\vDash\varphi\rightarrow\neg\psi$$ holds. A formula $$\theta(x_1\ldots x_n)$$ is said to be completable (in T) iff there is a complete formula $$\varphi(x_1\ldots x_n)$$ with $$T\vDash\varphi\rightarrow\theta$$. If $$\theta(x_1\ldots x_n)$$ is not completable it is said to be incompletable.

A theory $$T$$ is said to be atomic iff every formula of $$\mathcal{L}$$ which is consistent with $$T$$ is completable in $$T$$. A model $$\mathfrak{U}$$ if is said to be an atomic model iff every $$n$$-tuple $$a_1,\ldots,a_n\in A$$ satisfies a complete formula in $$\mathrm{Th}(\mathfrak{U})$$.

$$T$$ is complete iff for every sentence $$\phi$$ either $$T \models \phi$$ or $$T \models \lnot \phi$$. This requirement only applies to sentences, i.e., closed formulas: if $$\phi = \phi(x_1, \ldots, x_n)$$ has free variables then $$\phi$$ might hold under some interpretations of $$\phi$$ but not under others. E.g., consider $$\phi(x, y) = x < y$$ in the theory of a dense linear order without endpoints.

The formula $$x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$$ is complete in the sense of the definition you give, because for any $$\phi$$, $$T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$$ iff $$T \models \phi(c_1, \ldots, c_n)$$. Now, $$\phi(c_1, \ldots, c_n)$$ is a sentence, so if $$T$$ is complete either $$T \models \phi(c_1, \ldots, c_n)$$ or $$T \models \lnot\phi(c_1, \ldots, c_n)$$ and so either $$T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$$ or $$T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \lnot\phi$$.

If every element of a model $$\mathfrak{A}$$ is a constant, then every $$n$$-tuple, $$a_1, \ldots a_n$$ satisfies the formula $$x_1\equiv a_1\wedge x_1\equiv a_2\wedge\cdots\wedge x_n\equiv a_n$$, which we have just shown is complete. Hence such an $$\mathfrak{A}$$ is an atomic model according to the definition you have quoted.

• Where did you get $$T \models x_{0} \equiv c_{0}∧x_{1}\equiv c_{1}∧⋯∧x_{n}\equiv c_{n}\rightarrow\phi$$ iff$$T \models\phi\left( c_{1},…,c_{n}\right)$$ string? Mustn't T be bounded for that? There are nothing in example about boundness. Or, maybe we can derive it from condition of complete T? Feb 7, 2021 at 16:11
• This is true in any structure. An assignment to the $x_i$ either makes $x_1 \equiv c_1 \land \ldots x_n \equiv c_n$ false or it assigns $c_i$ to $x_i$ for each $i$ reducing the implication to $\phi(c_1, \ldots, c_n)$. Feb 7, 2021 at 16:16
• But what about formulas, not sentences? Or we have, that every model of $T \cup x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ will also realize $\phi\left( c_1,...c_n \right)$ and, as consequence, $\phi\left( x_1,...x_n \right)$ ? Feb 7, 2021 at 16:49
• But no, then it's consequence won't be just $\phi\left( x_1,...x_n \right)$, it will be $\exists x_1...\exists x_n\phi\left( x_1,...x_n \right)$, won't it? Feb 7, 2021 at 16:52
• No, this just follows from the definition of $\models$: $T \models \alpha$ means that in every model $\mathfrak{M}$ of $T$ we have $\mathfrak{M} \models \alpha$ and this means that $\alpha$ holds under every assignment of elements of $M$ to the free variables of $\alpha$. Feb 7, 2021 at 21:14

I changed my answer to myself:

Thus $$T$$ is complete, therefore it's consistent therefore it has model. So, it's sufficient to show, that $$T \cup x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$$ also will be consistent (and from that complete).

Suppose it isn't true.

Then, $$T \models \forall x_0...\forall x_n \neg(x_0 \equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n)$$ or $$T \models \forall x_0...\forall x_n (\neg x_0 \equiv c_0\vee \neg x_1\equiv c_1 \vee \cdots\vee \neg x_n\equiv c_n)$$

Interpretation $$I^{\mathfrak{A}}_{}$$ of $$c_{i} := t_{2i}$$, as a term from signature in the model $$\mathfrak{A} = $$ is $$I^{\mathfrak{A}}_{}[c_i] = a_{i}$$.

Some labeling $$\gamma_{j}$$ of $$x_{i} := t_{1i}$$, as a term, in model $$\mathfrak{A} =$$ is $$t^{\mathfrak{A}}_{1i}[\gamma_{j}] = \gamma_{j} x_{i} = a_{1i}$$.

So, thus in the last formula with all variables bounded, it's must be true on all labelings $$\gamma_{j}$$ and on all of our variables.

But for all $$i \in {0,...,n}$$, $$x_{i}$$ takes all values from $$A$$, and eventually it takes $$a_i$$ in some labeling $$\gamma_{j}$$.

The resulting contradiction completes the proof.

• In 1. if $T$ models every sentence, then $T$ is inconsistent - a much stronger property than completeness, which mens that $T$ either proves or disproves every sentence. Likewise in 2. it should read "or the same for $\lnot \phi$. An example of a formula that is not complete is $\phi = x < y \lor x = y$ in the theory of a dense linear order without endpoints: with $\psi = x < y$, $DLO \not\models \phi \to \psi$ and $DLO \not\models \phi \to \lnot \psi$. Feb 8, 2021 at 15:12
• So, where did you explain, in your answer, steps with that "forall" symbols? Feb 8, 2021 at 15:49
• There are no forall symbols involved in the question or in my answer. It is certainly not the case that $T \cup x_0 \equiv c_0 \land\ldots \models \phi$ implies that $T \models \forall x_0, \ldots \phi(x_0, \ldots)$. Feb 8, 2021 at 15:54
• I take $\mathfrak{M} \models \phi$ where $\phi$ has free variables to mean that $\mathfrak{M} \models \phi$ for every interpretation of the free variables in $\mathfrak{M}$. You can state this in terms of a universal quantifier. You are not manipulating the quantifiers correctly. Feb 8, 2021 at 16:08
• No, the correct definition is this: A theory $T$ is complete if for every sentence $\varphi$, $T\models \varphi$ or $T\models \lnot \varphi$. Sentences have no free variables. Rob Arthan's answer is correct. I was going to post another answer to help you understand it, but I can already tell that that won't be productive. Feb 9, 2021 at 22:24