YES

Problem:
 f(x) -> f(f(x))
 g(x) -> f(x)
 f(x) -> g(x)

Proof:
 Church Rosser Transformation Processor:
  strict:
   
  weak:
   
  critical peaks: 2
   f(f(x)) <-0|[]- f(x) -2|[]-> g(x)
   g(x) <-2|[]- f(x) -0|[]-> f(f(x))
  Shift Processor lab=left (dd) (force):
   strict:
    f(x) -> f(f(x))
    f(x) -> g(x)
    g(x) -> f(x)
    f(x) -> f(f(x))
    f(x) -> g(x)
    f(x) -> f(f(x))
    g(x) -> f(x)
    f(x) -> f(f(x))
   weak:
    f(x) -> f(f(x))
    g(x) -> f(x)
    f(x) -> g(x)
   Shift Processor (ldh) lab=left (force):
    strict:
     
    weak:
     
    Rule Labeling Processor:
     strict:
      
     weak:
      
     rule labeling (right):
      f(x) -> f(f(x)): 0
      g(x) -> f(x): 0
      f(x) -> g(x): 1
     Rule Labeling Processor:
      strict:
       
      weak:
       
      rule labeling (left):
       f(x) -> f(f(x)): 0
       g(x) -> f(x): 0
       f(x) -> g(x): 0
      Decreasing Processor:
       The following diagrams are decreasing:
       peak:
        f(f(x)) <-0|[==0,==1,=>0]- f(x) -2|[==0,==1,?=1]-> g(x)
       joins (1):
        
        g(x) -1|[==0,==1,=>0]-> f(x) -0|[==0,==1,=>0]-> f(f(x))
       peak:
        g(x) <-2|[==0,==1,=?1]- f(x) -0|[==0,==1,>=0]-> f(f(x))
       joins (1):
        g(x) -1|[==0,==1,>=0]-> f(x) -0|[==0,==1,>=0]-> f(f(x))
        
       Qed