YES

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

Proof:
 DP Processor:
  DPs:
   b#(f(b(x,z)),y) -> b#(y,z)
   b#(f(b(x,z)),y) -> b#(z,b(y,z))
   c#(f(f(c(x,a(),z))),a(),y) -> b#(a(),z)
   c#(f(f(c(x,a(),z))),a(),y) -> b#(y,f(b(a(),z)))
  TRS:
   b(f(b(x,z)),y) -> f(f(f(b(z,b(y,z)))))
   c(f(f(c(x,a(),z))),a(),y) -> b(y,f(b(a(),z)))
   b(b(c(b(a(),a()),a(),z),f(a())),y) -> z
  Usable Rule Processor:
   DPs:
    b#(f(b(x,z)),y) -> b#(y,z)
    b#(f(b(x,z)),y) -> b#(z,b(y,z))
    c#(f(f(c(x,a(),z))),a(),y) -> b#(a(),z)
    c#(f(f(c(x,a(),z))),a(),y) -> b#(y,f(b(a(),z)))
   TRS:
    b(f(b(x,z)),y) -> f(f(f(b(z,b(y,z)))))
    b(b(c(b(a(),a()),a(),z),f(a())),y) -> z
   Matrix Interpretation Processor: dim=3
    
    usable rules:
     b(f(b(x,z)),y) -> f(f(f(b(z,b(y,z)))))
     b(b(c(b(a(),a()),a(),z),f(a())),y) -> z
    interpretation:
     [c#](x0, x1, x2) = [1 0 0]x0 + [0 1 0]x1 + [1 0 0]x2 + [1],
     
     [b#](x0, x1) = [1 0 0]x0 + [1 0 0]x1,
     
                       [0 0 0]     [1 0 0]     [0]
     [c](x0, x1, x2) = [0 1 0]x1 + [0 1 1]x2 + [1]
                       [0 1 0]     [1 1 1]     [1],
     
           [0]
     [a] = [1]
           [0],
     
               [0 1 0]     [1]
     [f](x0) = [0 0 1]x0 + [0]
               [0 0 0]     [0],
     
                        [0 0 0]  
     [b](x0, x1) = x0 + [1 1 0]x1
                        [0 0 0]  
    orientation:
     b#(f(b(x,z)),y) = [0 1 0]x + [1 0 0]y + [1 1 0]z + [1] >= [1 0 0]y + [1 0 0]z = b#(y,z)
     
     b#(f(b(x,z)),y) = [0 1 0]x + [1 0 0]y + [1 1 0]z + [1] >= [1 0 0]y + [1 0 0]z = b#(z,b(y,z))
     
     c#(f(f(c(x,a(),z))),a(),y) = [1 0 0]y + [1 1 1]z + [5] >= [1 0 0]z = b#(a(),z)
     
     c#(f(f(c(x,a(),z))),a(),y) = [1 0 0]y + [1 1 1]z + [5] >= [1 0 0]y + [1 1 0]z + [2] = b#(y,f(b(a(),z)))
     
                      [0 1 0]    [0 0 0]    [1 1 0]    [1]    [1]                       
     b(f(b(x,z)),y) = [0 0 1]x + [1 1 0]y + [0 0 0]z + [0] >= [0] = f(f(f(b(z,b(y,z)))))
                      [0 0 0]    [0 0 0]    [0 0 0]    [0]    [0]                       
     
                                          [0 0 0]    [1 0 0]    [0]         
     b(b(c(b(a(),a()),a(),z),f(a())),y) = [1 1 0]y + [0 1 1]z + [4] >= z = z
                                          [0 0 0]    [1 1 1]    [2]         
    problem:
     DPs:
      
     TRS:
      b(f(b(x,z)),y) -> f(f(f(b(z,b(y,z)))))
      b(b(c(b(a(),a()),a(),z),f(a())),y) -> z
    Qed