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 0 1]  
    [s](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 0]  
    [a](x0) = [0 1 0]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [c](x0) = [0 0 0]x0 + [1]
              [0 0 0]     [0],
    
              [1 0 0]     [0]
    [f](x0) = [0 0 0]x0 + [1]
              [0 1 1]     [0],
    
              [1 0 0]     [0]
    [n](x0) = [0 0 0]x0 + [1]
              [0 1 1]     [1]
   orientation:
            [1 0 0]     [0]    [1 0 0]     [0]                 
    f(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(n(c(n(x1))))
            [0 1 1]     [0]    [0 0 0]     [0]                 
    
               [1 0 0]     [0]    [1 0 0]     [0]                 
    f(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = c(a(n(f(x1))))
               [0 0 0]     [1]    [0 0 0]     [0]                 
    
               [1 0 0]     [0]    [1 0 0]     [0]        
    a(n(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = c(x1)
               [0 0 0]     [0]    [0 0 0]     [0]        
    
               [1 0 0]     [0]    [1 0 0]     [0]        
    c(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = c(x1)
               [0 0 0]     [0]    [0 0 0]     [0]        
    
               [1 1 1]     [1]    [1 1 1]                
    s(n(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = s(s(f(x1)))
               [0 0 0]     [0]    [0 0 0]                
    
               [1 0 0]     [0]    [1 0 0]     [0]           
    f(n(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = n(f(x1))
               [0 1 1]     [2]    [0 1 1]     [2]           
   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))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(c) = w(n) = w(a) = w(f) = 0
    status function:
     st(c) = st(n) = st(a) = st(f) = [0]
    precedence:
     f > a > c ~ n
    problem:
     
    Qed