YES

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

Proof:
 DP Processor:
  DPs:
   c#(c(z,y,a()),a(),a()) -> b#(z,y)
   f#(c(x,y,z)) -> b#(y,z)
   f#(c(x,y,z)) -> f#(b(y,z))
   f#(c(x,y,z)) -> c#(z,f(b(y,z)),a())
   b#(z,b(c(a(),y,a()),f(f(x)))) -> c#(y,a(),z)
   b#(z,b(c(a(),y,a()),f(f(x)))) -> c#(c(y,a(),z),z,x)
  TRS:
   c(c(z,y,a()),a(),a()) -> b(z,y)
   f(c(x,y,z)) -> c(z,f(b(y,z)),a())
   b(z,b(c(a(),y,a()),f(f(x)))) -> c(c(y,a(),z),z,x)
  Matrix Interpretation Processor: dim=2
   
   usable rules:
    c(c(z,y,a()),a(),a()) -> b(z,y)
    b(z,b(c(a(),y,a()),f(f(x)))) -> c(c(y,a(),z),z,x)
   interpretation:
    [f#](x0) = [1 0]x0,
    
    [b#](x0, x1) = [0 1]x1,
    
    [c#](x0, x1, x2) = [0 1]x0 + [1],
    
              [0 1]     [1]
    [f](x0) = [1 0]x0 + [0],
    
                  [0 2]     [0 2]  
    [b](x0, x1) = [0 2]x0 + [0 2]x1,
    
                      [0 2]     [0 2]     [0 2]     [2]
    [c](x0, x1, x2) = [0 2]x0 + [0 1]x1 + [0 0]x2 + [0],
    
          [0]
    [a] = [0]
   orientation:
    c#(c(z,y,a()),a(),a()) = [0 1]y + [0 2]z + [1] >= [0 1]y = b#(z,y)
    
    f#(c(x,y,z)) = [0 2]x + [0 2]y + [0 2]z + [2] >= [0 1]z = b#(y,z)
    
    f#(c(x,y,z)) = [0 2]x + [0 2]y + [0 2]z + [2] >= [0 2]y + [0 2]z = f#(b(y,z))
    
    f#(c(x,y,z)) = [0 2]x + [0 2]y + [0 2]z + [2] >= [0 1]z + [1] = c#(z,f(b(y,z)),a())
    
    b#(z,b(c(a(),y,a()),f(f(x)))) = [0 2]x + [0 2]y + [2] >= [0 1]y + [1] = c#(y,a(),z)
    
    b#(z,b(c(a(),y,a()),f(f(x)))) = [0 2]x + [0 2]y + [2] >= [0 2]y + [1] = c#(c(y,a(),z),z,x)
    
                            [0 2]    [0 4]    [2]    [0 2]    [0 2]          
    c(c(z,y,a()),a(),a()) = [0 2]y + [0 4]z + [0] >= [0 2]y + [0 2]z = b(z,y)
    
                  [0 2]    [0 1]    [0 0]    [1]    [0 4]    [0 6]    [2]                     
    f(c(x,y,z)) = [0 2]x + [0 2]y + [0 2]z + [2] >= [0 2]y + [0 4]z + [0] = c(z,f(b(y,z)),a())
    
                                   [0 4]    [0 4]    [0 2]    [4]    [0 2]    [0 4]    [0 2]    [2]                    
    b(z,b(c(a(),y,a()),f(f(x)))) = [0 4]x + [0 4]y + [0 2]z + [4] >= [0 0]x + [0 4]y + [0 1]z + [0] = c(c(y,a(),z),z,x)
   problem:
    DPs:
     
    TRS:
     c(c(z,y,a()),a(),a()) -> b(z,y)
     f(c(x,y,z)) -> c(z,f(b(y,z)),a())
     b(z,b(c(a(),y,a()),f(f(x)))) -> c(c(y,a(),z),z,x)
   Qed