YES

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

Proof:
 DP Processor:
  DPs:
   b#(b(y,z),c(a(),a(),a())) -> c#(z,y,z)
   b#(b(y,z),c(a(),a(),a())) -> f#(c(z,y,z))
   c#(y,x,f(z)) -> b#(z,x)
   c#(y,x,f(z)) -> f#(b(z,x))
   c#(y,x,f(z)) -> b#(f(b(z,x)),z)
  TRS:
   b(b(y,z),c(a(),a(),a())) -> f(c(z,y,z))
   f(b(b(a(),z),c(a(),x,y))) -> z
   c(y,x,f(z)) -> b(f(b(z,x)),z)
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [f#](x0) = [0],
    
    [c#](x0, x1, x2) = [2 0]x0 + [2 0]x1 + [2 3]x2 + [1],
    
    [b#](x0, x1) = [2 0]x0 + [2 0]x1,
    
              [0 1]  
    [f](x0) = [2 0]x0,
    
                      [0 0]     [2 2]  
    [c](x0, x1, x2) = [1 0]x1 + [0 2]x2,
    
          [1]
    [a] = [0],
    
                  [1 0]     [2 2]  
    [b](x0, x1) = [2 0]x0 + [0 0]x1
   orientation:
    b#(b(y,z),c(a(),a(),a())) = [2 0]y + [4 4]z + [4] >= [2 0]y + [4 3]z + [1] = c#(z,y,z)
    
    b#(b(y,z),c(a(),a(),a())) = [2 0]y + [4 4]z + [4] >= [0] = f#(c(z,y,z))
    
    c#(y,x,f(z)) = [2 0]x + [2 0]y + [6 2]z + [1] >= [2 0]x + [2 0]z = b#(z,x)
    
    c#(y,x,f(z)) = [2 0]x + [2 0]y + [6 2]z + [1] >= [0] = f#(b(z,x))
    
    c#(y,x,f(z)) = [2 0]x + [2 0]y + [6 2]z + [1] >= [6 0]z = b#(f(b(z,x)),z)
    
                               [1 0]    [2 2]    [6]    [1 0]    [0 2]               
    b(b(y,z),c(a(),a(),a())) = [2 0]y + [4 4]z + [0] >= [0 0]y + [4 4]z = f(c(z,y,z))
    
                                [0 0]    [0  0 ]    [4 4]    [2]         
    f(b(b(a(),z),c(a(),x,y))) = [4 0]x + [8  16]y + [4 4]z + [2] >= z = z
    
                  [0 0]    [4 2]     [4 2]                  
    c(y,x,f(z)) = [1 0]x + [4 0]z >= [4 0]z = b(f(b(z,x)),z)
   problem:
    DPs:
     
    TRS:
     b(b(y,z),c(a(),a(),a())) -> f(c(z,y,z))
     f(b(b(a(),z),c(a(),x,y))) -> z
     c(y,x,f(z)) -> b(f(b(z,x)),z)
   Qed