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
   
   usable rules:
    c(c(b(c(x)))) -> b(a(0(),c(x)))
    c(c(x)) -> b(c(b(c(x))))
    a(0(),x) -> c(c(x))
   interpretation:
    [a#](x0, x1) = [3 0 1]x0 + [2 0 2]x1,
    
    [c#](x0) = [2 0 0]x0 + [1],
    
                  [0 2 0]     [3 2 1]     [2]
    [a](x0, x1) = [0 0 0]x0 + [3 0 1]x1 + [1]
                  [1 1 0]     [2 0 2]     [0],
    
          [1]
    [0] = [1]
          [1],
    
              [0 1 0]     [0]
    [b](x0) = [0 0 0]x0 + [1]
              [0 0 1]     [2],
    
              [1 0 1]     [1]
    [c](x0) = [1 0 1]x0 + [0]
              [2 0 0]     [0]
   orientation:
    c#(c(b(c(x)))) = [6 0 2]x + [7] >= [6 0 2]x + [6] = a#(0(),c(x))
    
    c#(c(x)) = [2 0 2]x + [3] >= [2 0 2]x + [1] = c#(b(c(x)))
    
    a#(0(),x) = [2 0 2]x + [4] >= [2 0 0]x + [1] = c#(x)
    
    a#(0(),x) = [2 0 2]x + [4] >= [2 0 2]x + [3] = c#(c(x))
    
                    [5 0 3]    [4]    [5 0 3]    [4]                 
    c(c(b(c(x)))) = [5 0 3]x + [3] >= [0 0 0]x + [1] = b(a(0(),c(x)))
                    [6 0 2]    [6]    [6 0 2]    [6]                 
    
              [3 0 1]    [2]    [3 0 1]    [2]                
    c(c(x)) = [3 0 1]x + [1] >= [0 0 0]x + [1] = b(c(b(c(x))))
              [2 0 2]    [2]    [2 0 2]    [2]                
    
               [3 2 1]    [4]    [3 0 1]    [2]          
    a(0(),x) = [3 0 1]x + [1] >= [3 0 1]x + [1] = c(c(x))
               [2 0 2]    [2]    [2 0 2]    [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