YES

Problem:
 f(a()) -> f(f(a()))
 a() -> b()
 f(x) -> f(b())

Proof:
 Church Rosser Transformation Processor (no redundant rules):
  strict:
   f(x) -> f(b())
   a() -> b()
   f(a()) -> f(f(a()))
  weak:
   
  critical peaks: 3
   f(b()) <-0|[]- f(a()) -2|[]-> f(f(a()))
   f(b()) <-1|0[]- f(a()) -2|[]-> f(f(a()))
   f(f(a())) <-2|[]- f(a()) -0|[]-> f(b())
  Closedness Processor (*upside-parallel*):
   
   Qed