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))))
  Arctic Interpretation Processor:
   dimension: 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,
    
    [c](x0) = 8x0 + 0,
    
    [a](x0) = 8x0,
    
    [b](x0) = -8x0 + 0
   orientation:
    a#(c(x1)) = 8x1 + 0 >= x1 = a#(x1)
    
    a#(c(x1)) = 8x1 + 0 >= 8x1 = a#(a(x1))
    
    a(b(x1)) = x1 + 8 >= x1 = x1
    
    a(b(x1)) = x1 + 8 >= x1 + 0 = b(c(x1))
    
    a(c(x1)) = 16x1 + 8 >= 16x1 + 8 = c(b(a(a(x1))))
   problem:
    DPs:
     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))))
   Restore Modifier:
    DPs:
     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) = 1/2x0,
      
      [c](x0) = 2x0 + 2,
      
      [a](x0) = 2x0,
      
      [b](x0) = 1/2x0 + 1
     orientation:
      a#(c(x1)) = x1 + 1 >= x1 = a#(a(x1))
      
      a(b(x1)) = x1 + 2 >= x1 = x1
      
      a(b(x1)) = x1 + 2 >= x1 + 2 = b(c(x1))
      
      a(c(x1)) = 4x1 + 4 >= 4x1 + 4 = 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