YES

1 decompositions
 #1 -----------
   1: f(g(x1,a(),b())) -> x1
   2: g(f(h(c(),d())),x2,x3) -> h(k1(x2),k2(x3))
   3: k1(a()) -> c()
   4: k2(b()) -> d()

@Knuth and Bendix' criterion
--- R
   1: f(g(x1,a(),b())) -> x1
   2: g(f(h(c(),d())),x2,x3) -> h(k1(x2),k2(x3))
   3: k1(a()) -> c()
   4: k2(b()) -> d()

--- S
   1: f(g(x1,a(),b())) -> x1
   2: g(f(h(c(),d())),x2,x3) -> h(k1(x2),k2(x3))
   3: k1(a()) -> c()
   4: k2(b()) -> d()


NOTE: input TRS is reduced

original is
   1: f(g(x1,a(),b())) -> x1
   2: g(f(h(c(),d())),x2,x3) -> h(k1(x2),k2(x3))
   3: k1(a()) -> c()
   4: k2(b()) -> d()
   5: f(h(k1(a()),k2(b()))) -> f(h(c(),d()))
   6: f(h(c(),k2(b()))) -> f(h(c(),d()))
   7: f(h(k1(a()),d())) -> f(h(c(),d()))

reduced to
   1: f(g(x1,a(),b())) -> x1
   2: g(f(h(c(),d())),x2,x3) -> h(k1(x2),k2(x3))
   3: k1(a()) -> c()
   4: k2(b()) -> d()