YES

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

Proof:
 DP Processor:
  DPs:
   f#(c(c(a(),y,a()),b(x,z),a())) -> f#(a())
   f#(c(c(a(),y,a()),b(x,z),a())) -> c#(f(a()),z,z)
   f#(c(c(a(),y,a()),b(x,z),a())) -> f#(c(f(a()),z,z))
   f#(b(b(x,f(y)),z)) -> f#(a())
   f#(b(b(x,f(y)),z)) -> f#(b(b(f(a()),y),y))
   f#(b(b(x,f(y)),z)) -> c#(z,x,f(b(b(f(a()),y),y)))
  TRS:
   f(c(c(a(),y,a()),b(x,z),a())) -> b(y,f(c(f(a()),z,z)))
   f(b(b(x,f(y)),z)) -> c(z,x,f(b(b(f(a()),y),y)))
   c(b(a(),a()),b(y,z),x) -> b(a(),b(z,z))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c#](x0, x1, x2) = 4x0,
    
    [f#](x0) = 2x0,
    
    [f](x0) = 2x0 + 1,
    
    [b](x0, x1) = 2x0 + 3x1 + 1,
    
    [c](x0, x1, x2) = x0 + 5x1,
    
    [a] = 0
   orientation:
    f#(c(c(a(),y,a()),b(x,z),a())) = 20x + 10y + 30z + 10 >= 0 = f#(a())
    
    f#(c(c(a(),y,a()),b(x,z),a())) = 20x + 10y + 30z + 10 >= 4 = c#(f(a()),z,z)
    
    f#(c(c(a(),y,a()),b(x,z),a())) = 20x + 10y + 30z + 10 >= 10z + 2 = f#(c(f(a()),z,z))
    
    f#(b(b(x,f(y)),z)) = 8x + 24y + 6z + 18 >= 0 = f#(a())
    
    f#(b(b(x,f(y)),z)) = 8x + 24y + 6z + 18 >= 18y + 14 = f#(b(b(f(a()),y),y))
    
    f#(b(b(x,f(y)),z)) = 8x + 24y + 6z + 18 >= 4z = c#(z,x,f(b(b(f(a()),y),y)))
    
    f(c(c(a(),y,a()),b(x,z),a())) = 20x + 10y + 30z + 11 >= 2y + 30z + 10 = b(y,f(c(f(a()),z,z)))
    
    f(b(b(x,f(y)),z)) = 8x + 24y + 6z + 19 >= 5x + z = c(z,x,f(b(b(f(a()),y),y)))
    
    c(b(a(),a()),b(y,z),x) = 10y + 15z + 6 >= 15z + 4 = b(a(),b(z,z))
   problem:
    DPs:
     
    TRS:
     f(c(c(a(),y,a()),b(x,z),a())) -> b(y,f(c(f(a()),z,z)))
     f(b(b(x,f(y)),z)) -> c(z,x,f(b(b(f(a()),y),y)))
     c(b(a(),a()),b(y,z),x) -> b(a(),b(z,z))
   Qed