Is it possible that a theory A is not interpretable in a theory B, and B is not interpretable in A too?

Question

Suppose A and B are first order theories(probably, the restriction to first order theory is not essential, but for simplicity).

When one theory has finite model,the question is easy.

If one has the model of less (finite)cardinality than the other (at least cardinality case), then the one is interpretable in the other.

Then, A is interpretable in B, or B is interpretable in A.

However, when both theories have no finite model, the answer to the question is not obvious(at least to me).

In that case, the 2 situation are common.

1. One is interpretable in the other, but not vice versa (e.g. PA is interpretable in ZF).

2. Two theories are biinterpretable (e.g. ZF and ZFC are biinterpretable).

Then, is it possible that A is not interpretable in B, and B is not interpretable in A too?

My own answer to the question is Yes. It seems that Dan Willard's self-verifying theory(SVT) and PA is an example of such case. Because SVT can prove the consistency of PA, then SVT is not interpretable in PA, and PA's diagonal argument is not representable in SVT, then PA is not interpretable in SVT.

But, I'm not so familiar to Dan Willard's theory, so I'm not sure.

Then, is my own answer right? or are there more example of such theories?