YES

Problem:
 f(f(x,y),z) -> f(x,f(y,z))
 f(1(),x) -> x

Proof:
 Church Rosser Transformation Processor (kb):
  f(f(x,y),z) -> f(x,f(y,z))
  f(1(),x) -> x
  critical peaks: joinable
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [1] = 1,
     
     [f](x0, x1) = 3x0 + x1 + 4
    orientation:
     f(f(x,y),z) = 9x + 3y + z + 16 >= 3x + 3y + z + 8 = f(x,f(y,z))
     
     f(1(),x) = x + 7 >= x = x
    problem:
     
    Qed