YES

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

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