YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [c](x0) = [0 0 0]x0
             [0 0 0]  ,
   
             [1 0 0]     [0]
   [s](x0) = [0 0 0]x0 + [1]
             [1 1 0]     [0],
   
             [1 0 0]  
   [f](x0) = [0 1 0]x0
             [0 0 0]  ,
   
             [1 1 0]  
   [n](x0) = [0 1 0]x0
             [0 0 1]  
  orientation:
           [1 0 0]      [1 0 0]                
   f(x1) = [0 1 0]x1 >= [0 0 0]x1 = n(c(c(x1)))
           [0 0 0]      [0 0 0]                
   
              [1 0 0]      [1 0 0]                
   c(f(x1)) = [0 0 0]x1 >= [0 0 0]x1 = f(c(c(x1)))
              [0 0 0]      [0 0 0]                
   
              [1 0 0]      [1 0 0]          
   c(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(x1)
              [0 0 0]      [0 0 0]          
   
              [1 0 0]     [1]    [1 0 0]     [0]              
   n(s(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = f(s(s(x1)))
              [1 1 0]     [0]    [0 0 0]     [0]              
   
              [1 1 0]      [1 1 0]             
   n(f(x1)) = [0 1 0]x1 >= [0 1 0]x1 = f(n(x1))
              [0 0 0]      [0 0 0]             
  problem:
   f(x1) -> n(c(c(x1)))
   c(f(x1)) -> f(c(c(x1)))
   c(c(x1)) -> c(x1)
   n(f(x1)) -> f(n(x1))
  String Reversal Processor:
   f(x1) -> c(c(n(x1)))
   f(c(x1)) -> c(c(f(x1)))
   c(c(x1)) -> c(x1)
   f(n(x1)) -> n(f(x1))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(n) = w(f) = 3
     w(c) = 0
    status function:
     st(n) = st(c) = st(f) = [0]
    precedence:
     f > n ~ c
    problem:
     
    Qed