YES

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

Proof:
 DP Processor:
  DPs:
   b#(a(a(x1))) -> c#(x1)
   b#(a(a(x1))) -> b#(c(x1))
   b#(a(a(x1))) -> a#(b(c(x1)))
   c#(a(x1)) -> c#(x1)
   c#(a(x1)) -> a#(c(x1))
   b#(c(a(x1))) -> c#(x1)
   b#(c(a(x1))) -> b#(c(x1))
   b#(c(a(x1))) -> a#(b(c(x1)))
   c#(b(x1)) -> a#(x1)
   c#(b(x1)) -> b#(a(x1))
   a#(c(b(x1))) -> a#(x1)
   a#(c(b(x1))) -> b#(a(x1))
   a#(c(b(x1))) -> c#(b(a(x1)))
  TRS:
   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   b(c(a(x1))) -> a(b(c(x1)))
   c(b(x1)) -> b(a(x1))
   a(c(b(x1))) -> c(b(a(x1)))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a#](x0) = 4x0 + 1,
    
    [c#](x0) = 4x0 + 8,
    
    [b#](x0) = x0 + 3,
    
    [c](x0) = 4x0 + 7,
    
    [b](x0) = x0 + 1,
    
    [a](x0) = 4x0 + 8
   orientation:
    b#(a(a(x1))) = 16x1 + 43 >= 4x1 + 8 = c#(x1)
    
    b#(a(a(x1))) = 16x1 + 43 >= 4x1 + 10 = b#(c(x1))
    
    b#(a(a(x1))) = 16x1 + 43 >= 16x1 + 33 = a#(b(c(x1)))
    
    c#(a(x1)) = 16x1 + 40 >= 4x1 + 8 = c#(x1)
    
    c#(a(x1)) = 16x1 + 40 >= 16x1 + 29 = a#(c(x1))
    
    b#(c(a(x1))) = 16x1 + 42 >= 4x1 + 8 = c#(x1)
    
    b#(c(a(x1))) = 16x1 + 42 >= 4x1 + 10 = b#(c(x1))
    
    b#(c(a(x1))) = 16x1 + 42 >= 16x1 + 33 = a#(b(c(x1)))
    
    c#(b(x1)) = 4x1 + 12 >= 4x1 + 1 = a#(x1)
    
    c#(b(x1)) = 4x1 + 12 >= 4x1 + 11 = b#(a(x1))
    
    a#(c(b(x1))) = 16x1 + 45 >= 4x1 + 1 = a#(x1)
    
    a#(c(b(x1))) = 16x1 + 45 >= 4x1 + 11 = b#(a(x1))
    
    a#(c(b(x1))) = 16x1 + 45 >= 16x1 + 44 = c#(b(a(x1)))
    
    b(a(a(x1))) = 16x1 + 41 >= 16x1 + 40 = a(b(c(x1)))
    
    c(a(x1)) = 16x1 + 39 >= 16x1 + 36 = a(c(x1))
    
    b(c(a(x1))) = 16x1 + 40 >= 16x1 + 40 = a(b(c(x1)))
    
    c(b(x1)) = 4x1 + 11 >= 4x1 + 9 = b(a(x1))
    
    a(c(b(x1))) = 16x1 + 52 >= 16x1 + 43 = c(b(a(x1)))
   problem:
    DPs:
     
    TRS:
     b(a(a(x1))) -> a(b(c(x1)))
     c(a(x1)) -> a(c(x1))
     b(c(a(x1))) -> a(b(c(x1)))
     c(b(x1)) -> b(a(x1))
     a(c(b(x1))) -> c(b(a(x1)))
   Qed