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This may be more appropriate for MO but I thought I'd ask here first as it's just a question about logic (not my strong point at all but not research-level in itself). I'm going through Razborov's flag algebra paper (properly this time) and I'm struggling with what an open interpretation of logical theories actually is. It's defined on p.14 as follows:

Let $T_1$ and $T_2$ be two universal first order theories in respective languages $L_1$, $L_2$ which do not contain function symbols (they can contain predicate symbols and constants). Let $U(x)$ be an open formula in $L_2$ such that $T_2 \vdash U(c)$ for every constant $c\in L_2$. Let $I$ be a translation which sends each predicate symbol $P(x_1, \dots, x_r)\in L_1$ to an open formula $I(P)(x_1, \dots, x_r)\in L_2$ and each constant $c\in L_1$ to a constant $I(c) \in L_2$. Extend $I$ to open formulas in $L_1$ by declaring it commutes with logical connectives. The pair $(U, I)$ is called an open interpretation of $T_1$ in $T_2$ if for every axiom $\forall x_1, \dots, x_r A(x_1, \dots, x_r) $ of $T_1$, we have $$T_2 \vdash \forall x_1, \dots, x_r \left[\left(U(x_1)\wedge \dots \wedge U(x_r) \right)\implies I(A) (x_1, \dots, x_r) \right]$$

So the interpretation of every axiom (in this case some formula consisting of relations and constants) of $T_1$ is an open formula in $T_2$ and it is provable in $T_2$ that $U(x_1)\wedge \dots \wedge U(x_r)$ implies $I(A)(x_1, \dots, x_r)$ for any $x_1, \dots, x_r$.

I cannot get my head around this, although I follow the logical terminology. How do people think of such interpretations? Can I roughly think of this as "$U$ implies the interpretations of the axioms of $T_1$, giving an "interpreted" copy of $T_1$ living inside $T_2$"?

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Alex Saad
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  • We can say that the "predicate" $U(x)$ restricts the domain $D$ of the interpetation (i.e. the "universe" in which individual variables and quantifiers range) to the "subset" $U^D$ containing only those elements of $D$ such that $U(x)$ holds of them. – Mauro ALLEGRANZA Mar 12 '15 at 16:15
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    A silly example : consider a theory about the domain $P$ of $Philosophers$ and some axiom like $\forall x Bald(x)$, stating that (in the "intended" interpretation) : "all philosophers are bald". We can "restrict" this theory to the sub-domain of Greek philosophers with a suitable predicate $Greek(x)$ and rewrite the axiom as follows : $\forall x (Greek(x) \to Bald(x))$. – Mauro ALLEGRANZA Mar 12 '15 at 16:18
  • A silly but excellent example! This has really helped me. So for instance if my domain is "graphs" (an application of flag algebras) with an axiom "$\forall x T(x)$ with the interpretation "$x$ contains a triangle", then we can restrict to the sub-domain of graphs containing a 4-cycle with the predicate $C_4 (x)$ for "$x$ contains a 4-cycle". So in this restricted domain if a graph contains a 4-cycle it also contains a triangle. Right? – Alex Saad Mar 12 '15 at 16:36
  • More or less ... With non-silly examples, we have to prove the "interpretability". I mean : if $\forall x T(x)$ is an axiom (a "good" one) we expect that it is true, and this means that "every graph contains a triangle". If so, obviously also "a graph with a $4$-cycle contains a triangle". What we need is to "map" the theory $T_1$ into $T_2$ with a suitable $U(x)$ such that for any axiom $\forall x \phi(x)$ of $T_1$ we are able to prove in $T_2$ that the "interpreted" axiom holds, i.e. that $T_2 \vdash \forall x [U(x) \to \phi^*(x)]$. – Mauro ALLEGRANZA Mar 12 '15 at 16:53
  • Okay, this is starting to make a lot of sense! Of course, I should perhaps have set my domain as "graphs containing a triangle" but regardless I'm beginning to see how this works and how it can be used to force combinatorial structures to have certain properties. Thanks so much! – Alex Saad Mar 12 '15 at 17:10
  • The answers to http://math.stackexchange.com/questions/1090437/how-to-prove-that-gödels-incompleteness-theorems-apply-to-zfc may also be of interest. – Rob Arthan Mar 12 '15 at 17:50
  • @RobArthan thanks for this link, this is also very helpful! – Alex Saad Mar 12 '15 at 18:21

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