YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(c(x1))) -> a#(b(x1))
   a#(b(c(x1))) -> a#(a(b(x1)))
  TRS:
   a(b(x1)) -> x1
   a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    a(b(x1)) -> x1
    a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
   interpretation:
    [a#](x0) = 1/2x0,
    
    [c](x0) = 2x0 + 1/2,
    
    [a](x0) = 2x0,
    
    [b](x0) = 1/2x0
   orientation:
    a#(b(c(x1))) = 1/2x1 + 1/8 >= 1/4x1 = a#(b(x1))
    
    a#(b(c(x1))) = 1/2x1 + 1/8 >= 1/2x1 = a#(a(b(x1)))
    
    a(b(x1)) = x1 >= x1 = x1
    
    a(b(c(x1))) = 2x1 + 1/2 >= 2x1 + 1/2 = b(c(b(c(a(a(b(x1)))))))
   problem:
    DPs:
     
    TRS:
     a(b(x1)) -> x1
     a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
   Qed