YES

Problem:
 a(b(x1)) -> b(b(a(x1)))
 c(b(x1)) -> b(c(c(x1)))

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