YES

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

Proof:
 DP Processor:
  DPs:
   a#(c(x1)) -> b#(x1)
   a#(x1) -> b#(x1)
   a#(x1) -> b#(b(x1))
   a#(x1) -> b#(b(b(x1)))
   b#(c(b(x1))) -> a#(c(x1))
  TRS:
   a(c(x1)) -> c(b(x1))
   a(x1) -> b(b(b(x1)))
   b(c(b(x1))) -> a(c(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b#](x0) = 2x0 + 2,
    
    [a#](x0) = 2x0 + 7,
    
    [b](x0) = x0 + 1,
    
    [a](x0) = x0 + 3,
    
    [c](x0) = 3x0 + 2
   orientation:
    a#(c(x1)) = 6x1 + 11 >= 2x1 + 2 = b#(x1)
    
    a#(x1) = 2x1 + 7 >= 2x1 + 2 = b#(x1)
    
    a#(x1) = 2x1 + 7 >= 2x1 + 4 = b#(b(x1))
    
    a#(x1) = 2x1 + 7 >= 2x1 + 6 = b#(b(b(x1)))
    
    b#(c(b(x1))) = 6x1 + 12 >= 6x1 + 11 = a#(c(x1))
    
    a(c(x1)) = 3x1 + 5 >= 3x1 + 5 = c(b(x1))
    
    a(x1) = x1 + 3 >= x1 + 3 = b(b(b(x1)))
    
    b(c(b(x1))) = 3x1 + 6 >= 3x1 + 5 = a(c(x1))
   problem:
    DPs:
     
    TRS:
     a(c(x1)) -> c(b(x1))
     a(x1) -> b(b(b(x1)))
     b(c(b(x1))) -> a(c(x1))
   Qed