YES

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

Proof:
 DP Processor:
  DPs:
   b#(y,z) -> c#(y,z,z)
   b#(y,z) -> c#(c(y,z,z),a(),a())
   c#(f(z),f(c(a(),x,a())),y) -> c#(z,y,a())
   c#(f(z),f(c(a(),x,a())),y) -> b#(x,z)
   c#(f(z),f(c(a(),x,a())),y) -> c#(f(b(x,z)),c(z,y,a()),a())
  TRS:
   b(y,z) -> f(c(c(y,z,z),a(),a()))
   b(b(z,y),a()) -> z
   c(f(z),f(c(a(),x,a())),y) -> c(f(b(x,z)),c(z,y,a()),a())
  Matrix Interpretation Processor: dim=3
   
   usable rules:
    b(y,z) -> f(c(c(y,z,z),a(),a()))
    b(b(z,y),a()) -> z
    c(f(z),f(c(a(),x,a())),y) -> c(f(b(x,z)),c(z,y,a()),a())
   interpretation:
    [c#](x0, x1, x2) = [0 2 0]x0 + [0 1 1]x1 + [0 1 1]x2,
    
    [b#](x0, x1) = [0 2 0]x0 + [0 2 2]x1 + [3],
    
              [2 0 0]  
    [f](x0) = [0 1 1]x0
              [2 0 0]  ,
    
          [0]
    [a] = [1]
          [0],
    
                      [0 2 0]     [0 1 1]     [2 0 1]  
    [c](x0, x1, x2) = [0 0 0]x0 + [0 0 0]x1 + [0 0 0]x2
                      [0 2 0]     [0 0 1]     [0 0 2]  ,
    
                  [1 1 0]     [0 0 0]     [2]
    [b](x0, x1) = [0 1 0]x0 + [0 0 1]x1 + [0]
                  [0 0 1]     [0 0 0]     [2]
   orientation:
    b#(y,z) = [0 2 0]y + [0 2 2]z + [3] >= [0 2 0]y + [0 2 2]z = c#(y,z,z)
    
    b#(y,z) = [0 2 0]y + [0 2 2]z + [3] >= [2] = c#(c(y,z,z),a(),a())
    
    c#(f(z),f(c(a(),x,a())),y) = [0 2 3]x + [0 1 1]y + [0 2 2]z + [6] >= [0 1 1]y + [0 2 0]z + [1] = c#(z,y,a())
    
    c#(f(z),f(c(a(),x,a())),y) = [0 2 3]x + [0 1 1]y + [0 2 2]z + [6] >= [0 2 0]x + [0 2 2]z + [3] = b#(x,z)
    
    c#(f(z),f(c(a(),x,a())),y) = [0 2 3]x + [0 1 1]y + [0 2 2]z + [6] >= [0 2 2]x + [0 0 1]y + [0 2 2]z + [5] = c#(f(b(x,z)),c(z,y,a()),a())
    
             [1 1 0]    [0 0 0]    [2]    [2]                         
    b(y,z) = [0 1 0]y + [0 0 1]z + [0] >= [0] = f(c(c(y,z,z),a(),a()))
             [0 0 1]    [0 0 0]    [2]    [2]                         
    
                    [0 0 1]    [1 2 0]    [4]         
    b(b(z,y),a()) = [0 0 1]y + [0 1 0]z + [0] >= z = z
                    [0 0 0]    [0 0 1]    [4]         
    
                                [0 2 3]    [2 0 1]    [0 2 2]    [6]    [0 2 2]    [0 0 1]    [0 2 2]    [4]                              
    c(f(z),f(c(a(),x,a())),y) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] >= [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] = c(f(b(x,z)),c(z,y,a()),a())
                                [0 2 2]    [0 0 2]    [0 2 2]    [4]    [0 2 2]    [0 0 1]    [0 2 2]    [4]                              
   problem:
    DPs:
     
    TRS:
     b(y,z) -> f(c(c(y,z,z),a(),a()))
     b(b(z,y),a()) -> z
     c(f(z),f(c(a(),x,a())),y) -> c(f(b(x,z)),c(z,y,a()),a())
   Qed