TITLE: Why is it undecidable whether two finite-state transducers are equivalent?
QUESTION [10 upvotes]: According to the Wikipedia page on finite-state transducers, it is undecidable whether two finite-state transducers are equivalent.  I find this result striking, since it is decidable whether two finite-state automata are equivalent to one another.
Unfortunately, Wikipedia doesn't provide any citations that provide a justification for this result.  Does anyone know of a proof of this result?  Alternatively, if the article is incorrect, does someone know an algorithm that could be used to show equivalence?
Thanks!

REPLY [6 votes]: To complement on Brian's answer, you might want to look at The equivalence problem of multitape finite automata, where the authors give an algorithm for deciding whether two deterministic transducers are equivalent. The key is that it is decidable if two automata are equivalent with regard to multiplicities (i.e. how many times the word is accepted). 
This is very closely related to probabilistic automata, of which the equivalence algorithm was given in A polynomial-time algorithm for the equivalence of probabilistic automata. In this setting the question is whether two automata accept the word with the same probability.
You can generalize that for some weighted automatons, without loosing decidability (see here for nice slides with examples). Thus, you could transform deterministic (multitape) automatons into weighted version such that word is accepted if and only if it has (in some semiring) weight 1 (one), and afterwards just use the previous algorithm. However, please note that this is rather fragile result, even adding $\varepsilon$-edges would make this undecidable.
Hope this helps ;-)