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:
 DP Processor:
  DPs:
   f#(x1) -> n#(a(x1))
   f#(x1) -> c#(n(a(x1)))
   f#(x1) -> n#(c(n(a(x1))))
   c#(f(x1)) -> c#(x1)
   c#(f(x1)) -> n#(a(c(x1)))
   c#(f(x1)) -> f#(n(a(c(x1))))
   n#(a(x1)) -> c#(x1)
   n#(s(x1)) -> f#(s(s(x1)))
   n#(f(x1)) -> n#(x1)
   n#(f(x1)) -> f#(n(x1))
  TRS:
   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))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [c#](x0) = [1 0]x0 + [1],
    
    [n#](x0) = [1 1]x0 + [2],
    
    [f#](x0) = [1 0]x0 + [4],
    
              [1 0]     [0]
    [s](x0) = [0 0]x0 + [4],
    
              [1 0]  
    [c](x0) = [0 0]x0,
    
              [1 1]  
    [n](x0) = [0 1]x0,
    
              [1 0]  
    [a](x0) = [0 0]x0,
    
                   [4]
    [f](x0) = x0 + [0]
   orientation:
    f#(x1) = [1 0]x1 + [4] >= [1 0]x1 + [2] = n#(a(x1))
    
    f#(x1) = [1 0]x1 + [4] >= [1 0]x1 + [1] = c#(n(a(x1)))
    
    f#(x1) = [1 0]x1 + [4] >= [1 0]x1 + [2] = n#(c(n(a(x1))))
    
    c#(f(x1)) = [1 0]x1 + [5] >= [1 0]x1 + [1] = c#(x1)
    
    c#(f(x1)) = [1 0]x1 + [5] >= [1 0]x1 + [2] = n#(a(c(x1)))
    
    c#(f(x1)) = [1 0]x1 + [5] >= [1 0]x1 + [4] = f#(n(a(c(x1))))
    
    n#(a(x1)) = [1 0]x1 + [2] >= [1 0]x1 + [1] = c#(x1)
    
    n#(s(x1)) = [1 0]x1 + [6] >= [1 0]x1 + [4] = f#(s(s(x1)))
    
    n#(f(x1)) = [1 1]x1 + [6] >= [1 1]x1 + [2] = n#(x1)
    
    n#(f(x1)) = [1 1]x1 + [6] >= [1 1]x1 + [4] = f#(n(x1))
    
                 [4]    [1 0]                   
    f(x1) = x1 + [0] >= [0 0]x1 = n(c(n(a(x1))))
    
               [1 0]     [4]    [1 0]     [4]                 
    c(f(x1)) = [0 0]x1 + [0] >= [0 0]x1 + [0] = f(n(a(c(x1))))
    
               [1 0]      [1 0]          
    n(a(x1)) = [0 0]x1 >= [0 0]x1 = c(x1)
    
               [1 0]      [1 0]          
    c(c(x1)) = [0 0]x1 >= [0 0]x1 = c(x1)
    
               [1 0]     [4]    [1 0]     [4]              
    n(s(x1)) = [0 0]x1 + [4] >= [0 0]x1 + [4] = f(s(s(x1)))
    
               [1 1]     [4]    [1 1]     [4]           
    n(f(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = f(n(x1))
   problem:
    DPs:
     
    TRS:
     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))
   Qed