YES

Problem:
 r(r(x1)) -> s(r(x1))
 r(s(x1)) -> s(r(x1))
 r(n(x1)) -> s(r(x1))
 r(b(x1)) -> u(s(b(x1)))
 r(u(x1)) -> u(r(x1))
 s(u(x1)) -> u(s(x1))
 n(u(x1)) -> u(n(x1))
 t(r(u(x1))) -> t(c(r(x1)))
 t(s(u(x1))) -> t(c(r(x1)))
 t(n(u(x1))) -> t(c(r(x1)))
 c(u(x1)) -> u(c(x1))
 c(s(x1)) -> s(c(x1))
 c(r(x1)) -> r(c(x1))
 c(n(x1)) -> n(c(x1))
 c(n(x1)) -> n(x1)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [u](x0) = 4x0 + 1,
   
   [s](x0) = x0,
   
   [b](x0) = 4x0 + 4,
   
   [t](x0) = x0 + 4,
   
   [r](x0) = 4x0 + 1,
   
   [c](x0) = x0,
   
   [n](x0) = 4x0
  orientation:
   r(r(x1)) = 16x1 + 5 >= 4x1 + 1 = s(r(x1))
   
   r(s(x1)) = 4x1 + 1 >= 4x1 + 1 = s(r(x1))
   
   r(n(x1)) = 16x1 + 1 >= 4x1 + 1 = s(r(x1))
   
   r(b(x1)) = 16x1 + 17 >= 16x1 + 17 = u(s(b(x1)))
   
   r(u(x1)) = 16x1 + 5 >= 16x1 + 5 = u(r(x1))
   
   s(u(x1)) = 4x1 + 1 >= 4x1 + 1 = u(s(x1))
   
   n(u(x1)) = 16x1 + 4 >= 16x1 + 1 = u(n(x1))
   
   t(r(u(x1))) = 16x1 + 9 >= 4x1 + 5 = t(c(r(x1)))
   
   t(s(u(x1))) = 4x1 + 5 >= 4x1 + 5 = t(c(r(x1)))
   
   t(n(u(x1))) = 16x1 + 8 >= 4x1 + 5 = t(c(r(x1)))
   
   c(u(x1)) = 4x1 + 1 >= 4x1 + 1 = u(c(x1))
   
   c(s(x1)) = x1 >= x1 = s(c(x1))
   
   c(r(x1)) = 4x1 + 1 >= 4x1 + 1 = r(c(x1))
   
   c(n(x1)) = 4x1 >= 4x1 = n(c(x1))
   
   c(n(x1)) = 4x1 >= 4x1 = n(x1)
  problem:
   r(s(x1)) -> s(r(x1))
   r(n(x1)) -> s(r(x1))
   r(b(x1)) -> u(s(b(x1)))
   r(u(x1)) -> u(r(x1))
   s(u(x1)) -> u(s(x1))
   t(s(u(x1))) -> t(c(r(x1)))
   c(u(x1)) -> u(c(x1))
   c(s(x1)) -> s(c(x1))
   c(r(x1)) -> r(c(x1))
   c(n(x1)) -> n(c(x1))
   c(n(x1)) -> n(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [u](x0) = 2x0 + 10,
    
    [s](x0) = x0,
    
    [b](x0) = 8x0 + 3,
    
    [t](x0) = x0 + 5,
    
    [r](x0) = 2x0 + 10,
    
    [c](x0) = x0,
    
    [n](x0) = x0 + 2
   orientation:
    r(s(x1)) = 2x1 + 10 >= 2x1 + 10 = s(r(x1))
    
    r(n(x1)) = 2x1 + 14 >= 2x1 + 10 = s(r(x1))
    
    r(b(x1)) = 16x1 + 16 >= 16x1 + 16 = u(s(b(x1)))
    
    r(u(x1)) = 4x1 + 30 >= 4x1 + 30 = u(r(x1))
    
    s(u(x1)) = 2x1 + 10 >= 2x1 + 10 = u(s(x1))
    
    t(s(u(x1))) = 2x1 + 15 >= 2x1 + 15 = t(c(r(x1)))
    
    c(u(x1)) = 2x1 + 10 >= 2x1 + 10 = u(c(x1))
    
    c(s(x1)) = x1 >= x1 = s(c(x1))
    
    c(r(x1)) = 2x1 + 10 >= 2x1 + 10 = r(c(x1))
    
    c(n(x1)) = x1 + 2 >= x1 + 2 = n(c(x1))
    
    c(n(x1)) = x1 + 2 >= x1 + 2 = n(x1)
   problem:
    r(s(x1)) -> s(r(x1))
    r(b(x1)) -> u(s(b(x1)))
    r(u(x1)) -> u(r(x1))
    s(u(x1)) -> u(s(x1))
    t(s(u(x1))) -> t(c(r(x1)))
    c(u(x1)) -> u(c(x1))
    c(s(x1)) -> s(c(x1))
    c(r(x1)) -> r(c(x1))
    c(n(x1)) -> n(c(x1))
    c(n(x1)) -> n(x1)
   String Reversal Processor:
    s(r(x1)) -> r(s(x1))
    b(r(x1)) -> b(s(u(x1)))
    u(r(x1)) -> r(u(x1))
    u(s(x1)) -> s(u(x1))
    u(s(t(x1))) -> r(c(t(x1)))
    u(c(x1)) -> c(u(x1))
    s(c(x1)) -> c(s(x1))
    r(c(x1)) -> c(r(x1))
    n(c(x1)) -> c(n(x1))
    n(c(x1)) -> n(x1)
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                [1 0 0]  
      [u](x0) = [0 1 0]x0
                [1 1 1]  ,
      
                [1 1 0]  
      [s](x0) = [0 0 0]x0
                [0 0 1]  ,
      
                [1 1 1]  
      [b](x0) = [0 0 0]x0
                [0 0 0]  ,
      
                [1 0 0]     [0]
      [t](x0) = [0 0 0]x0 + [0]
                [1 0 1]     [1],
      
                [1 0 0]     [0]
      [r](x0) = [0 1 0]x0 + [0]
                [1 1 1]     [1],
      
                [1 0 0]  
      [c](x0) = [0 1 0]x0
                [0 0 0]  ,
      
                [1 0 0]     [0]
      [n](x0) = [0 0 0]x0 + [1]
                [0 0 0]     [0]
     orientation:
                 [1 1 0]     [0]    [1 1 0]     [0]           
      s(r(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = r(s(x1))
                 [1 1 1]     [1]    [1 1 1]     [1]           
      
                 [2 2 1]     [1]    [2 2 1]                
      b(r(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = b(s(u(x1)))
                 [0 0 0]     [0]    [0 0 0]                
      
                 [1 0 0]     [0]    [1 0 0]     [0]           
      u(r(x1)) = [0 1 0]x1 + [0] >= [0 1 0]x1 + [0] = r(u(x1))
                 [2 2 1]     [1]    [2 2 1]     [1]           
      
                 [1 1 0]      [1 1 0]             
      u(s(x1)) = [0 0 0]x1 >= [0 0 0]x1 = s(u(x1))
                 [1 1 1]      [1 1 1]             
      
                    [1 0 0]     [0]    [1 0 0]     [0]              
      u(s(t(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = r(c(t(x1)))
                    [2 0 1]     [1]    [1 0 0]     [1]              
      
                 [1 0 0]      [1 0 0]             
      u(c(x1)) = [0 1 0]x1 >= [0 1 0]x1 = c(u(x1))
                 [1 1 0]      [0 0 0]             
      
                 [1 1 0]      [1 1 0]             
      s(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(s(x1))
                 [0 0 0]      [0 0 0]             
      
                 [1 0 0]     [0]    [1 0 0]             
      r(c(x1)) = [0 1 0]x1 + [0] >= [0 1 0]x1 = c(r(x1))
                 [1 1 0]     [1]    [0 0 0]             
      
                 [1 0 0]     [0]    [1 0 0]     [0]           
      n(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = c(n(x1))
                 [0 0 0]     [0]    [0 0 0]     [0]           
      
                 [1 0 0]     [0]    [1 0 0]     [0]        
      n(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = n(x1)
                 [0 0 0]     [0]    [0 0 0]     [0]        
     problem:
      s(r(x1)) -> r(s(x1))
      u(r(x1)) -> r(u(x1))
      u(s(x1)) -> s(u(x1))
      u(s(t(x1))) -> r(c(t(x1)))
      u(c(x1)) -> c(u(x1))
      s(c(x1)) -> c(s(x1))
      r(c(x1)) -> c(r(x1))
      n(c(x1)) -> c(n(x1))
      n(c(x1)) -> n(x1)
     KBO Processor:
      weight function:
       w0 = 1
       w(c) = w(t) = w(u) = w(s) = w(r) = 1
       w(n) = 0
      precedence:
       n > t > u > s > r > c
      problem:
       
      Qed