YES

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

Proof:
 DP Processor:
  DPs:
   a#(a(x1)) -> b#(a(x1))
   a#(a(x1)) -> a#(b(a(x1)))
   b#(a(b(x1))) -> a#(x1)
   b#(a(b(x1))) -> a#(c(a(x1)))
  TRS:
   a(a(x1)) -> a(b(a(x1)))
   b(a(b(x1))) -> a(c(a(x1)))
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    a(a(x1)) -> a(b(a(x1)))
    b(a(b(x1))) -> a(c(a(x1)))
   interpretation:
    [b#](x0) = 1x0 + 0,
    
    [a#](x0) = 2x0 + 0,
    
    [c](x0) = 0,
    
    [b](x0) = -5x0 + 6,
    
    [a](x0) = 7x0 + 7
   orientation:
    a#(a(x1)) = 9x1 + 9 >= 8x1 + 8 = b#(a(x1))
    
    a#(a(x1)) = 9x1 + 9 >= 4x1 + 8 = a#(b(a(x1)))
    
    b#(a(b(x1))) = 3x1 + 14 >= 2x1 + 0 = a#(x1)
    
    b#(a(b(x1))) = 3x1 + 14 >= 2 = a#(c(a(x1)))
    
    a(a(x1)) = 14x1 + 14 >= 9x1 + 13 = a(b(a(x1)))
    
    b(a(b(x1))) = -3x1 + 8 >= 7 = a(c(a(x1)))
   problem:
    DPs:
     
    TRS:
     a(a(x1)) -> a(b(a(x1)))
     b(a(b(x1))) -> a(c(a(x1)))
   Qed