0 equivalence relation.
We show that indeed appears to be a natural example of a Π 1 0
equivalence relation is obtained in this way. A slight extension of the
every Π 1 f E argument yields a function such that the corresponding
equivalence relation f 0 f is Π -complete. We show that can in fact
be chosen polynomial time computable. 1 0 E For each Π equivalence
relation , there is a computable sequence of cofinite 1 [2] E ( E E t
relations ( ) contained in with = such that from we can compute t t ( t t
∈ E E E , we have and a strong index for the complement of t t t +1
⊇ ( [2] E , t [0 ] is transitive. (2) t ∩ 0 Proposition . For
each (3) equivalence relation E, there is a computable binary 3.1 Π 1
, namely, for each x, y, we have function f such that E E f = x E y f f . =
x y ↔ Proof. E f x, n x ( f x, n Let ( ) x r E , x,n max( ) be as in
(2). We define ( ) by recursion on . Let ) = t t ( f r, n , r < x r for
the least such that otherwise. ( ) if x, ∈ (
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