YES

Problem:
 f(x1) -> n(c(n(a(x1))))
 c(f(x1)) -> f(n(a(c(x1))))
 n(a(x1)) -> c(x1)
 c(c(x1)) -> c(x1)
 n(s(x1)) -> f(s(s(x1)))
 n(f(x1)) -> f(n(x1))

Proof:
 String Reversal Processor:
  f(x1) -> a(n(c(n(x1))))
  f(c(x1)) -> c(a(n(f(x1))))
  a(n(x1)) -> c(x1)
  c(c(x1)) -> c(x1)
  s(n(x1)) -> s(s(f(x1)))
  f(n(x1)) -> n(f(x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 1 0]     [0]
    [s](x0) = [0 0 0]x0 + [0]
              [0 1 1]     [1],
    
              [1 0 1]  
    [c](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 1]     [0]
    [n](x0) = [0 1 0]x0 + [1]
              [0 0 0]     [0],
    
              [1 0 0]  
    [a](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 1]  
    [f](x0) = [0 1 0]x0
              [0 0 0]  
   orientation:
            [1 0 1]      [1 0 1]                   
    f(x1) = [0 1 0]x1 >= [0 0 0]x1 = a(n(c(n(x1))))
            [0 0 0]      [0 0 0]                   
    
               [1 0 1]      [1 0 1]                   
    f(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(a(n(f(x1))))
               [0 0 0]      [0 0 0]                   
    
               [1 0 1]      [1 0 1]          
    a(n(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(x1)
               [0 0 0]      [0 0 0]          
    
               [1 0 1]      [1 0 1]          
    c(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(x1)
               [0 0 0]      [0 0 0]          
    
               [1 1 1]     [1]    [1 1 1]     [0]              
    s(n(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = s(s(f(x1)))
               [0 1 0]     [2]    [0 1 0]     [2]              
    
               [1 0 1]     [0]    [1 0 1]     [0]           
    f(n(x1)) = [0 1 0]x1 + [1] >= [0 1 0]x1 + [1] = n(f(x1))
               [0 0 0]     [0]    [0 0 0]     [0]           
   problem:
    f(x1) -> a(n(c(n(x1))))
    f(c(x1)) -> c(a(n(f(x1))))
    a(n(x1)) -> c(x1)
    c(c(x1)) -> c(x1)
    f(n(x1)) -> n(f(x1))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0  -&]  
     [c](x0) = [-& -&]x0,
     
               [0  -&]  
     [n](x0) = [0  -&]x0,
     
               [0  0 ]  
     [a](x0) = [-& -&]x0,
     
               [1  -&]  
     [f](x0) = [0  1 ]x0
    orientation:
             [1  -&]      [0  -&]                   
     f(x1) = [0  1 ]x1 >= [-& -&]x1 = a(n(c(n(x1))))
     
                [1  -&]      [1  -&]                   
     f(c(x1)) = [0  -&]x1 >= [-& -&]x1 = c(a(n(f(x1))))
     
                [0  -&]      [0  -&]          
     a(n(x1)) = [-& -&]x1 >= [-& -&]x1 = c(x1)
     
                [0  -&]      [0  -&]          
     c(c(x1)) = [-& -&]x1 >= [-& -&]x1 = c(x1)
     
                [1  -&]      [1  -&]             
     f(n(x1)) = [1  -&]x1 >= [1  -&]x1 = n(f(x1))
    problem:
     f(c(x1)) -> c(a(n(f(x1))))
     a(n(x1)) -> c(x1)
     c(c(x1)) -> c(x1)
     f(n(x1)) -> n(f(x1))
    String Reversal Processor:
     c(f(x1)) -> f(n(a(c(x1))))
     n(a(x1)) -> c(x1)
     c(c(x1)) -> c(x1)
     n(f(x1)) -> f(n(x1))
     Matrix Interpretation Processor: dim=3
      
      interpretation:
                 [1 0 0]  
       [c](x0) = [0 0 0]x0
                 [0 0 1]  ,
       
                 [1 0 1]  
       [n](x0) = [0 0 0]x0
                 [0 1 1]  ,
       
                 [1 0 0]  
       [a](x0) = [0 0 1]x0
                 [0 0 0]  ,
       
                 [1 0 0]     [0]
       [f](x0) = [0 0 0]x0 + [0]
                 [0 1 1]     [1]
      orientation:
                  [1 0 0]     [0]    [1 0 0]     [0]                 
       c(f(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = f(n(a(c(x1))))
                  [0 1 1]     [1]    [0 0 1]     [1]                 
       
                  [1 0 0]      [1 0 0]          
       n(a(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(x1)
                  [0 0 1]      [0 0 1]          
       
                  [1 0 0]      [1 0 0]          
       c(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(x1)
                  [0 0 1]      [0 0 1]          
       
                  [1 1 1]     [1]    [1 0 1]     [0]           
       n(f(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = f(n(x1))
                  [0 1 1]     [1]    [0 1 1]     [1]           
      problem:
       c(f(x1)) -> f(n(a(c(x1))))
       n(a(x1)) -> c(x1)
       c(c(x1)) -> c(x1)
      String Reversal Processor:
       f(c(x1)) -> c(a(n(f(x1))))
       a(n(x1)) -> c(x1)
       c(c(x1)) -> c(x1)
       Matrix Interpretation Processor: dim=3
        
        interpretation:
                        [0]
         [c](x0) = x0 + [1]
                        [1],
         
                     
         [n](x0) = x0
                     ,
         
                        [0]
         [a](x0) = x0 + [1]
                        [1],
         
                   [1 0 1]     [0]
         [f](x0) = [0 1 1]x0 + [0]
                   [0 1 1]     [1]
        orientation:
                    [1 0 1]     [1]    [1 0 1]     [0]                 
         f(c(x1)) = [0 1 1]x1 + [2] >= [0 1 1]x1 + [2] = c(a(n(f(x1))))
                    [0 1 1]     [3]    [0 1 1]     [3]                 
         
                         [0]         [0]        
         a(n(x1)) = x1 + [1] >= x1 + [1] = c(x1)
                         [1]         [1]        
         
                         [0]         [0]        
         c(c(x1)) = x1 + [2] >= x1 + [1] = c(x1)
                         [2]         [1]        
        problem:
         a(n(x1)) -> c(x1)
         c(c(x1)) -> c(x1)
        Arctic Interpretation Processor:
         dimension: 3
         interpretation:
                    [1  -& 1 ]  
          [c](x0) = [2  0  2 ]x0
                    [-& 0  1 ]  ,
          
                    [1 0 0]  
          [n](x0) = [1 0 1]x0
                    [2 3 0]  ,
          
                    [0  0  -&]  
          [a](x0) = [1  1  0 ]x0
                    [2  0  0 ]  
         orientation:
                     [1 0 1]      [1  -& 1 ]          
          a(n(x1)) = [2 3 2]x1 >= [2  0  2 ]x1 = c(x1)
                     [3 3 2]      [-& 0  1 ]          
          
                     [2 1 2]      [1  -& 1 ]          
          c(c(x1)) = [3 2 3]x1 >= [2  0  2 ]x1 = c(x1)
                     [2 1 2]      [-& 0  1 ]          
         problem:
          a(n(x1)) -> c(x1)
         Arctic Interpretation Processor:
          dimension: 3
          interpretation:
                     [0  -& -&]  
           [c](x0) = [-& -& -&]x0
                     [-& -& -&]  ,
           
                     [1  0  1 ]  
           [n](x0) = [0  0  1 ]x0
                     [2  0  -&]  ,
           
                     [2  1  1 ]  
           [a](x0) = [1  1  -&]x0
                     [1  -& 1 ]  
          orientation:
                      [3 2 3]      [0  -& -&]          
           a(n(x1)) = [2 1 2]x1 >= [-& -& -&]x1 = c(x1)
                      [3 1 2]      [-& -& -&]          
          problem:
           
          Qed