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
   interpretation:
    [b#](x0) = -2x0 + 1,
    
    [a#](x0) = x0 + -6,
    
    [c](x0) = 1,
    
    [b](x0) = -1x0 + 4,
    
    [a](x0) = 4x0 + 5
   orientation:
    a#(a(x1)) = 4x1 + 5 >= 2x1 + 3 = b#(a(x1))
    
    a#(a(x1)) = 4x1 + 5 >= 3x1 + 4 = a#(b(a(x1)))
    
    b#(a(b(x1))) = 1x1 + 6 >= x1 + -6 = a#(x1)
    
    b#(a(b(x1))) = 1x1 + 6 >= 1 = a#(c(a(x1)))
    
    a(a(x1)) = 8x1 + 9 >= 7x1 + 8 = a(b(a(x1)))
    
    b(a(b(x1))) = 2x1 + 7 >= 5 = a(c(a(x1)))
   problem:
    DPs:
     
    TRS:
     a(a(x1)) -> a(b(a(x1)))
     b(a(b(x1))) -> a(c(a(x1)))
   Qed