Does synonymy seep down to the fragments of theories?

Question

IF we have a synonymous interpretation between two theories $T$ and $H$ that uses translation $\tau$ from the language of $T$ to the language of $H$. Then I'd expect that for a sentence $\mu$ in the language of $T$, $ T+ \mu$ (if consistent) would be also synonymous to $ H + \mu^\tau$, where $ \mu^\tau$ is the sentence resulting from translating $\mu$ to the language of $H$ using $\tau$. Now, can this work also for a fragment of $T$? that is, suppose that $S$ is a sub-theory of $T$ (i.e. all axioms of $S$ are true in $T$), then would the translation of the axioms of $S$ to the language of $H$ using $\tau$ also be synonymous with $S$?

Answer 1

It's a very nice question!

The answer is negative. For a counterexample, consider:

• Let $T$ be the theory of a partial order $\leq$, that is, a reflexive, transitive, antisymmetric relation.
• Let $H$ be the theory of a strict partial order $<$, that is, an irreflexive transitive relation.

These theories are bi-interpretable, since from any partial order $\leq$ we can define the associated strict order via $$a<b\quad\iff\quad (a\leq b)\text{ and }\neg(b\leq a)$$ and conversely from the strict order we define the $\leq$ relation via $$a\leq b\quad\iff\quad (a<b)\text{ or }(a=b).$$ If we iterate these translations, we land in each case back at the original order. So the two order relations have a common definitional expansion and so this is a definitional equivalence, which is a particularly strong form of being a synonymy (one for which the bi-interpretation maps are the identity function).

Now let $S$ be the theory of a partial preorder. This is a proper subtheory of $T$, where we omit the antisymmetry requirement. So now we might have distinct elements $a\neq b$ that are equivalent in the sense that $a\leq b\leq a$. If we follow the interpretation into the language of $<$, we get the usual strict order associated with a partial preorder.

But this translation is not a synonymy, since from the strict order of a preorder, we cannot always recover the original preorder. The reason is that different preorders can give rise to exactly the same strict order.

For example, consider the preference relations shown here:

While Horace finds the soft and hard boiled options incomparable, Hortencia finds them equivalent—she is indifferent between them—but they have exactly the same strict preference instances. At bottom, the issue is that the strict order cannot tell the difference between incomparability and indifference in a preorder.

Meanwhile, see my related question, Is the theory of a partial order bi-interpretable with the theory of a pre-order? That question is mostly-but-not-quite-fully answered, but the suggestion is that the theory of a preorder is likely not bi-interpretable with the theory of a partial order, and therefore not bi-interpretable with the theory of a strict order. The one-dimensional claim in the answer of Rodrigo Freire shows that they are not synonymous.