YES

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

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