YES

1 decompositions
 #0 -----------
   1: H(H(x)) -> K(x)
   2: H(K(x)) -> K(H(x))

@Jouannaud and Kirchner's criterion
--- R
   1: H(H(x)) -> K(x)
   2: H(K(x)) -> K(H(x))

--- S
   1: H(H(x)) -> K(x)
   2: H(K(x)) -> K(H(x))