YES

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

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