MAYBE

Problem:
 f(x,a(b(y))) -> f(c(d(x)),y)
 f(c(x),y) -> f(x,a(y))
 f(d(x),y) -> f(x,b(y))

Proof:
 Complexity Transformation Processor:
  strict:
   f(x,a(b(y))) -> f(c(d(x)),y)
   f(c(x),y) -> f(x,a(y))
   f(d(x),y) -> f(x,b(y))
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   max_matrix:
    1
    interpretation:
     [c](x0) = x0,
     
     [d](x0) = x0 + 1,
     
     [f](x0, x1) = x0 + x1,
     
     [a](x0) = x0,
     
     [b](x0) = x0
    orientation:
     f(x,a(b(y))) = x + y >= x + y + 1 = f(c(d(x)),y)
     
     f(c(x),y) = x + y >= x + y = f(x,a(y))
     
     f(d(x),y) = x + y + 1 >= x + y = f(x,b(y))
    problem:
     strict:
      f(x,a(b(y))) -> f(c(d(x)),y)
      f(c(x),y) -> f(x,a(y))
     weak:
      f(d(x),y) -> f(x,b(y))
    Matrix Interpretation Processor:
     dimension: 1
     max_matrix:
      1
      interpretation:
       [c](x0) = x0 + 1,
       
       [d](x0) = x0,
       
       [f](x0, x1) = x0 + x1,
       
       [a](x0) = x0,
       
       [b](x0) = x0
      orientation:
       f(x,a(b(y))) = x + y >= x + y + 1 = f(c(d(x)),y)
       
       f(c(x),y) = x + y + 1 >= x + y = f(x,a(y))
       
       f(d(x),y) = x + y >= x + y = f(x,b(y))
      problem:
       strict:
        f(x,a(b(y))) -> f(c(d(x)),y)
       weak:
        f(c(x),y) -> f(x,a(y))
        f(d(x),y) -> f(x,b(y))
      Open