YES

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

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