Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 23 / 3 / c / Solution

Here ${\forall }_2$ denotes universal-existential sentences, with quantifier-free matrix and possibly empty blocks. Add constants for $A$ and considerWe show that $\Sigma$ is consistent. Suppose a finite subset is inconsistent. Combine its diagram facts as $\delta (\mathbf{c})$ and its finitely many negative existential facts as $\neg \exists {\mathbf{y}}_j{\theta }_j(\mathbf{c},{\mathbf{y}}_j)$. Replacing the new constants by a variable tuple gives the consequenceThis is equivalent to a universal-existential sentence: move all the existential blocks to the front of the disjunction, using the usual nonempty-domain convention. If there are no negative existential facts, the consequence is simply $\forall \mathbf{x}\neg \delta$. By hypothesis $A$ satisfies the displayed consequence. Its named tuple satisfies $\delta$ but none of the existential alternatives, a contradiction.

A model $B$ of $\Sigma$ contains a diagram copy of $A$. Every existential formula false in $A$ is still false in $B$; every one true in $A$ is preserved by the embedding. ThusBoth preservation and reflection are required here; the atomic diagram alone would give only an ordinary embedding.