Is rhyming of two words a transitive property?

Let's suppose A rhymes with B and B rhymes with C. Does A always rhyme with C?

Except in the technical case of equivocal words (where a single word has multiple pronunciations). If B was an equivocal word, it could rhyme with both A and C, yet A and C would not rhyme. For example:

A = taxes
B = axes
C = taxis

In this case, B is the plural of axe to rhyme with A, and the plural of axis to rhyme with C, but A and C do not rhyme.

(Some might argue that this shouldn't count, because B is really two words, not one. Fair enough; I did mention this counterexample was a technicality that breaks the transitive property of rhyming. I still felt it was worth a mention.)

EDIT. Since I've been called on my counterexample in the comments below, I offer one more example to mull over:

A = dollop \ˈdä-ləp\
B = scallop \ˈskä-ləp, ˈska-ləp\
C = gallop \ˈga-ləp\

Most dictionaries list two pronunciations for scallop; one rhymes with A (dollop), the other rhymes with C (gallop).

Now, B isn't technically two words, but one - albeit with two pronunciations. Still, A rhymes with B, and B rhymes with C, but A doesn't rhyme with C.

So, as the O.P. asked, "Does A always rhyme with C?"

Yes. "A rhymes with B" is a relationship of sameness; that is, a certain part of A is the same as the corresponding part of B. The mathematical term for this is an "equivalence relation", which is always reflexive (everything rhymes with itself), symmetric (if A rhymes with B, then B rhymes with A) and transitive.

If you're talking about perfect rhyme, the answer is no. The pair obtain and remain are a perfect rhyme, as are retain and remain, but obtain and retain do not form a perfect rhyme, since they both end with -tain, and the vowels before -tain are different. If you want rhymes to be transitive, you have to allow identical rhymes (see link above) and say that a word rhymes with itself.