MAYBE

Problem:
 a(b(x1)) -> d(x1)
 b(a(x1)) -> a(b(x1))
 d(c(x1)) -> f(a(b(b(c(x1)))))
 d(f(x1)) -> f(a(b(x1)))
 a(f(x1)) -> a(x1)

Proof:
 RT Transformation Processor:
  strict:
   a(b(x1)) -> d(x1)
   b(a(x1)) -> a(b(x1))
   d(c(x1)) -> f(a(b(b(c(x1)))))
   d(f(x1)) -> f(a(b(x1)))
   a(f(x1)) -> a(x1)
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [f](x0) = x0 + 7,
    
    [c](x0) = x0 + 16,
    
    [d](x0) = x0 + 25,
    
    [a](x0) = x0,
    
    [b](x0) = x0 + 12
   orientation:
    a(b(x1)) = x1 + 12 >= x1 + 25 = d(x1)
    
    b(a(x1)) = x1 + 12 >= x1 + 12 = a(b(x1))
    
    d(c(x1)) = x1 + 41 >= x1 + 47 = f(a(b(b(c(x1)))))
    
    d(f(x1)) = x1 + 32 >= x1 + 19 = f(a(b(x1)))
    
    a(f(x1)) = x1 + 7 >= x1 = a(x1)
   problem:
    strict:
     a(b(x1)) -> d(x1)
     b(a(x1)) -> a(b(x1))
     d(c(x1)) -> f(a(b(b(c(x1)))))
    weak:
     d(f(x1)) -> f(a(b(x1)))
     a(f(x1)) -> a(x1)
   Matrix Interpretation Processor:
    dimension: 1
    interpretation:
     [f](x0) = x0,
     
     [c](x0) = x0,
     
     [d](x0) = x0 + 2,
     
     [a](x0) = x0 + 1,
     
     [b](x0) = x0
    orientation:
     a(b(x1)) = x1 + 1 >= x1 + 2 = d(x1)
     
     b(a(x1)) = x1 + 1 >= x1 + 1 = a(b(x1))
     
     d(c(x1)) = x1 + 2 >= x1 + 1 = f(a(b(b(c(x1)))))
     
     d(f(x1)) = x1 + 2 >= x1 + 1 = f(a(b(x1)))
     
     a(f(x1)) = x1 + 1 >= x1 + 1 = a(x1)
    problem:
     strict:
      a(b(x1)) -> d(x1)
      b(a(x1)) -> a(b(x1))
     weak:
      d(c(x1)) -> f(a(b(b(c(x1)))))
      d(f(x1)) -> f(a(b(x1)))
      a(f(x1)) -> a(x1)
    Matrix Interpretation Processor:
     dimension: 2
     interpretation:
                  
      [f](x0) = x0,
      
                [1 3]  
      [c](x0) = [0 0]x0,
      
                [1 2]     [0]
      [d](x0) = [0 1]x0 + [1],
      
                     [0]
      [a](x0) = x0 + [1],
      
                [1 2]  
      [b](x0) = [0 1]x0
     orientation:
                 [1 2]     [0]    [1 2]     [0]        
      a(b(x1)) = [0 1]x1 + [1] >= [0 1]x1 + [1] = d(x1)
      
                 [1 2]     [2]    [1 2]     [0]           
      b(a(x1)) = [0 1]x1 + [1] >= [0 1]x1 + [1] = a(b(x1))
      
                 [1 3]     [0]    [1 3]     [0]                    
      d(c(x1)) = [0 0]x1 + [1] >= [0 0]x1 + [1] = f(a(b(b(c(x1)))))
      
                 [1 2]     [0]    [1 2]     [0]              
      d(f(x1)) = [0 1]x1 + [1] >= [0 1]x1 + [1] = f(a(b(x1)))
      
                      [0]         [0]        
      a(f(x1)) = x1 + [1] >= x1 + [1] = a(x1)
     problem:
      strict:
       a(b(x1)) -> d(x1)
      weak:
       b(a(x1)) -> a(b(x1))
       d(c(x1)) -> f(a(b(b(c(x1)))))
       d(f(x1)) -> f(a(b(x1)))
       a(f(x1)) -> a(x1)
     Open