YES

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

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