YES

Problem:
 b(a(),f(b(b(z,y),a()))) -> z
 c(c(z,x,a()),a(),y) -> f(f(c(y,a(),f(c(z,y,x)))))
 f(f(c(a(),y,z))) -> b(y,b(z,z))

Proof:
 DP Processor:
  DPs:
   c#(c(z,x,a()),a(),y) -> c#(z,y,x)
   c#(c(z,x,a()),a(),y) -> f#(c(z,y,x))
   c#(c(z,x,a()),a(),y) -> c#(y,a(),f(c(z,y,x)))
   c#(c(z,x,a()),a(),y) -> f#(c(y,a(),f(c(z,y,x))))
   c#(c(z,x,a()),a(),y) -> f#(f(c(y,a(),f(c(z,y,x)))))
   f#(f(c(a(),y,z))) -> b#(z,z)
   f#(f(c(a(),y,z))) -> b#(y,b(z,z))
  TRS:
   b(a(),f(b(b(z,y),a()))) -> z
   c(c(z,x,a()),a(),y) -> f(f(c(y,a(),f(c(z,y,x)))))
   f(f(c(a(),y,z))) -> b(y,b(z,z))
  Matrix Interpretation Processor: dim=3
   
   usable rules:
    b(a(),f(b(b(z,y),a()))) -> z
    c(c(z,x,a()),a(),y) -> f(f(c(y,a(),f(c(z,y,x)))))
    f(f(c(a(),y,z))) -> b(y,b(z,z))
   interpretation:
    [f#](x0) = [1],
    
    [c#](x0, x1, x2) = [1 0 0]x0 + [1 0 0]x1 + [1 0 0]x2 + [1],
    
    [b#](x0, x1) = [0],
    
                      [1 0 0]     [1 1 0]     [0 0 1]     [1]
    [c](x0, x1, x2) = [0 0 0]x0 + [0 0 0]x1 + [0 0 0]x2 + [0]
                      [0 0 0]     [1 1 1]     [1 0 0]     [0],
    
              [0 1 0]  
    [f](x0) = [0 0 1]x0
              [0 0 1]  ,
    
                  [0 0 0]     [1 0 0]  
    [b](x0, x1) = [1 1 1]x0 + [1 0 0]x1
                  [0 1 1]     [1 0 0]  ,
    
          [0]
    [a] = [0]
          [0]
   orientation:
    c#(c(z,x,a()),a(),y) = [1 1 0]x + [1 0 0]y + [1 0 0]z + [2] >= [1 0 0]x + [1 0 0]y + [1 0 0]z + [1] = c#(z,y,x)
    
    c#(c(z,x,a()),a(),y) = [1 1 0]x + [1 0 0]y + [1 0 0]z + [2] >= [1] = f#(c(z,y,x))
    
    c#(c(z,x,a()),a(),y) = [1 1 0]x + [1 0 0]y + [1 0 0]z + [2] >= [1 0 0]y + [1] = c#(y,a(),f(c(z,y,x)))
    
    c#(c(z,x,a()),a(),y) = [1 1 0]x + [1 0 0]y + [1 0 0]z + [2] >= [1] = f#(c(y,a(),f(c(z,y,x))))
    
    c#(c(z,x,a()),a(),y) = [1 1 0]x + [1 0 0]y + [1 0 0]z + [2] >= [1] = f#(f(c(y,a(),f(c(z,y,x)))))
    
    f#(f(c(a(),y,z))) = [1] >= [0] = b#(z,z)
    
    f#(f(c(a(),y,z))) = [1] >= [0] = b#(y,b(z,z))
    
                              [3 0 0]    [1 2 2]          
    b(a(),f(b(b(z,y),a()))) = [3 0 0]y + [1 2 2]z >= z = z
                              [3 0 0]    [1 2 2]          
    
                          [1 1 0]    [0 0 1]    [1 0 0]    [2]    [0]                             
    c(c(z,x,a()),a(),y) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] >= [0] = f(f(c(y,a(),f(c(z,y,x)))))
                          [0 0 0]    [1 0 0]    [0 0 0]    [0]    [0]                             
    
                       [1 1 1]    [1 0 0]     [0 0 0]    [1 0 0]               
    f(f(c(a(),y,z))) = [1 1 1]y + [1 0 0]z >= [1 1 1]y + [1 0 0]z = b(y,b(z,z))
                       [1 1 1]    [1 0 0]     [0 1 1]    [1 0 0]               
   problem:
    DPs:
     
    TRS:
     b(a(),f(b(b(z,y),a()))) -> z
     c(c(z,x,a()),a(),y) -> f(f(c(y,a(),f(c(z,y,x)))))
     f(f(c(a(),y,z))) -> b(y,b(z,z))
   Qed