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