Krdytkyu

Let $T$ be a theory in first order logic over some language $L$. Let $mathfrak A$ be some structure over $L$ with $mathfrak A models T$ and with $A$ be its universe. Then consider every $a in A$ as a constant and look at the enriched language $L(A) = L cup A$ with the $L(A)$-structure $mathfrak A_A = (mathfrak A, a)_{ain A}$. A formula over $L(A)$ is called basic if it is an atomic formula. The set
$$
operatorname{Diag}(mathfrak A) = { varphi mbox{ is a basic $L(A)$-sentence } mid mathfrak A_A models varphi }
$$
is called the atomic diagram of $mathfrak A$.

A theory $T$ is called model-complete if every substructure relation between two models is actually an elementary embedding.

Then $T$ is model-complete if and only if for any $mathfrak A models T$ the theory $T cup operatorname{Diag}(mathfrak A)$ is complete.

These definitions are from A course in model theory by K.Tent/M.Zeigler.

I do not understand the quoted statement. For let $t_1 = t_2$ and $t_3 ne t_4$ be two atomic sentences for terms $t_1, t_2,t_3,t_4$ in $L(A)$. Then Set $varphi = (t_1 = t_2) land (t_3 ne t_4)$. Now suppose in the terms we have some constants from $A$. Then neither $varphi$ nor $neg varphi$ is in $T cup operatorname{Diag}(mathfrak A)$ as it is not in $operatorname{Diag}(mathfrak A)$ for it is not atomic, nor is it in $T$ as it is a statement over the enrichted language $L(A)$, but not over $L$. Could someone please explain the above statement (and what I oversee here...)?

Complete means for any sentence $varphi,$ either $Tvdash varphi$ or $Tvdash lnot varphi,$ not $varphiin T$ or $lnotvarphiin T.$

If $mathfrak A,mathfrak Bmodels T$ and $mathfrak Asubseteq mathfrak B,$ then both are models of $Tcup Diag(mathfrak A).$ If $Tcup Diag(mathfrak A)$ is complete, then they must agree on all $L(A)$-sentences, and hence the embedding is elementary. On the other hand if $Tcup Diag(mathfrak A)$ is not complete for some $mathfrak Amodels T,$ then we can find a $mathfrak Bmodels T$ with $mathfrak Asubseteq mathfrak B$ that differs from $mathfrak A$ on some $L(A)$-sentence (just let it be a model of the negation of some $varphi$ that is true in $mathfrak A$ but that $Tcup Diag(mathfrak A)$ does not decide). Hence this embedding is not elementary.

You have two errors here:

1. "$T$ is complete" means that for every sentence $varphi$, $Tmodels varphi$ or $Tmodels lnotvarphi$. It does not mean that $varphiin T$ or $lnot varphiin T$.




2. A basic formula is not just an atomic formula. It is an atomic formula or a negation of an atomic formula.

In your example, if $mathfrak{A}models varphi$, then $t_1 = t_2in text{Diag}(mathfrak{A})$ and $t_3neq t_4in text{Diag}(mathfrak{A})$, so $Tcup text{Diag}(mathfrak{A})models varphi$. Otherwise, if $mathfrak{A}notmodels varphi$, then either $t_1neq t_2in text{Diag}(mathfrak{A})$ or $t_3 = t_4in text{Diag}(mathfrak{A})$, and in either case $Tcup text{Diag}(mathfrak{A})models lnot varphi$.

It's an easy exercise to show that for any structure $mathfrak{A}$ and any quantifier-free $L(A)$-sentence $varphi$, either $text{Diag}(mathfrak{A})models varphi$ or $text{Diag}(mathfrak{A})models lnot varphi$. The interesting fact is that $T$ is model complete if and only if $Tcuptext{Diag}(mathfrak{A})$ decides the truth of all $L(A)$-formulas, even those with quantifiers. The answer by spaceisdarkgreen explains the easy proof of this fact.