YES

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

Proof:
 Church Rosser Transformation Processor:
  strict:
   f(g(g(x))) -> a()
   f(g(h(x))) -> b()
   f(h(g(x))) -> b()
   f(h(h(x))) -> c()
   g(x) -> h(x)
   a() -> b()
   b() -> c()
  weak:
   
  critical peaks: 4
   f(h(g(x))) <-4|0[]- f(g(g(x))) -0|[]-> a()
   f(g(h(x))) <-4|0,0[]- f(g(g(x))) -0|[]-> a()
   f(h(h(x))) <-4|0[]- f(g(h(x))) -1|[]-> b()
   f(h(h(x))) <-4|0,0[]- f(h(g(x))) -2|[]-> b()
  Closedness Processor (*strongly -- <=7 steps*):
   strict:
    
   weak:
    
   Qed