YES

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

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