Abstract. Two algebraic structures with the same universe are called term-equivalent if they have the same clone of term operations. We show that the problem of determining whether two finite algebras of finite similarity type are term-equivalent is complete for deterministic exponential time.When are two algebraic structures (presumably of different similarity types) considered to be "the same", for all intents and purposes? This is the notion of 'term-equivalence', a cornerstone of universal algebra.For example, the two-element Boolean algebra is generally thought of as the structure B = {0, 1}, ∧, ∨, ¬ , with the basic operations being conjunction, disjunction and negation. However, it is sometimes convenient to use instead the algebra B = {0, 1}, | whose only basic operation is the Sheffer stroke. Algebraically speaking, these two structures behave in exactly the same way. It is easy to transform B into B and back via the definitionsInformally, two algebras A and B are called term-equivalent if each of the basic operations of A can be built from the basic operations of B, and vice-versa. For a precise formulation, see Definition 3.3.Because of its fundamental role, it is natural to wonder about the computational complexity of determining term-equivalence. Specifically, given two finite algebras A and B of finite similarity type, are A and B termequivalent? There is a straightforward deterministic algorithm for solving this problem (Theorem 3.6), and its running time is exponential in the size of the input. In this paper we prove that this is the best possible: the termequivalence problem is complete for EXPTIME (Theorem 3.8). It follows from the Hierarchy Theorem of Hartmanis and Stearns [5] that the class EXPTIME is strictly larger than PTIME (the class of problems solvable