YES

Problem:
 a(x,y) -> b(x,b(0(),c(y)))
 c(b(y,c(x))) -> c(c(b(a(0(),0()),y)))
 b(y,0()) -> y

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