MAYBE

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

Proof:
 Complexity Transformation Processor:
  strict:
   a(a(b(x1))) -> b(a(b(x1)))
   b(a(x1)) -> a(b(b(x1)))
   b(c(a(x1))) -> c(c(a(a(b(x1)))))
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   max_matrix:
    1
    interpretation:
     [c](x0) = x0,
     
     [a](x0) = x0 + 1,
     
     [b](x0) = x0
    orientation:
     a(a(b(x1))) = x1 + 2 >= x1 + 1 = b(a(b(x1)))
     
     b(a(x1)) = x1 + 1 >= x1 + 1 = a(b(b(x1)))
     
     b(c(a(x1))) = x1 + 1 >= x1 + 2 = c(c(a(a(b(x1)))))
    problem:
     strict:
      b(a(x1)) -> a(b(b(x1)))
      b(c(a(x1))) -> c(c(a(a(b(x1)))))
     weak:
      a(a(b(x1))) -> b(a(b(x1)))
    Matrix Interpretation Processor:
     dimension: 3
     max_matrix:
      [1 1 0]
      [0 0 1]
      [0 0 1]
      interpretation:
                 [1 0 0]  
       [c](x0) = [0 0 1]x0
                 [0 0 0]  ,
       
                 [1 1 0]     [0]
       [a](x0) = [0 0 0]x0 + [0]
                 [0 0 1]     [1],
       
                 [1 1 0]  
       [b](x0) = [0 0 0]x0
                 [0 0 1]  
      orientation:
                  [1 1 0]     [0]    [1 1 0]     [0]              
       b(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(b(x1)))
                  [0 0 1]     [1]    [0 0 1]     [1]              
       
                     [1 1 1]     [1]    [1 1 0]                      
       b(c(a(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c(c(a(a(b(x1)))))
                     [0 0 0]     [0]    [0 0 0]                      
       
                     [1 1 0]     [0]    [1 1 0]     [0]              
       a(a(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(b(x1)))
                     [0 0 1]     [2]    [0 0 1]     [1]              
      problem:
       strict:
        b(a(x1)) -> a(b(b(x1)))
       weak:
        b(c(a(x1))) -> c(c(a(a(b(x1)))))
        a(a(b(x1))) -> b(a(b(x1)))
      Open