YES

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

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