YES

Problem:
 f(g(x)) -> h(x,x)
 g(a()) -> b()
 f(x) -> h(x,x)
 b() -> a()
 h(x,y) -> h(g(x),g(y))
 g(x) -> x
 a() -> b()

Proof:
 Church Rosser Transformation Processor:
  strict:
   
  weak:
   
  critical peaks: 7
   h(x47,x47) <-0|[]- f(g(x47)) -2|[]-> h(g(x47),g(x47))
   f(b()) <-1|0[]- f(g(a())) -0|[]-> h(a(),a())
   b() <-1|[]- g(a()) -5|[]-> a()
   h(g(x),g(x)) <-2|[]- f(g(x)) -0|[]-> h(x,x)
   f(x) <-5|0[]- f(g(x)) -0|[]-> h(x,x)
   a() <-5|[]- g(a()) -1|[]-> b()
   g(b()) <-6|0[]- g(a()) -1|[]-> b()
  Redundant Rules Transformation:
   f(x) -> h(x,x)
   g(x) -> x
   a() -> b()
   Church Rosser Transformation Processor (no redundant rules):
    strict:
     a() -> b()
     g(x) -> x
     f(x) -> h(x,x)
    weak:
     
    critical peaks: 0
    Closedness Processor (*feeble*):
     
     Qed