YES

   1: f(i(x1),g(a())) -> f(j(x1,x1),g(b()))
   2: b() -> a()
   3: i(x1) -> j(x1,x1)

@Jouannaud and Kirchner's criterion
--- R
   1: f(i(x1),g(a())) -> f(j(x1,x1),g(b()))
   2: b() -> a()
   3: i(x1) -> j(x1,x1)

--- S
   1: f(i(x1),g(a())) -> f(j(x1,x1),g(b()))
   2: b() -> a()
   3: i(x1) -> j(x1,x1)