YES

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

Proof:
 DP Processor:
  DPs:
   a#(c(x1)) -> a#(x1)
   a#(c(x1)) -> a#(a(x1))
   a#(c(x1)) -> b#(c(a(a(x1))))
  TRS:
   a(a(b(x1))) -> c(x1)
   a(c(x1)) -> b(c(a(a(x1))))
   b(c(x1)) -> x1
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b#](x0) = x0 + 7/2,
    
    [a#](x0) = 4x0 + 6,
    
    [c](x0) = 2x0 + 13/2,
    
    [a](x0) = 2x0 + 1,
    
    [b](x0) = 1/2x0 + 4
   orientation:
    a#(c(x1)) = 8x1 + 32 >= 4x1 + 6 = a#(x1)
    
    a#(c(x1)) = 8x1 + 32 >= 8x1 + 10 = a#(a(x1))
    
    a#(c(x1)) = 8x1 + 32 >= 8x1 + 16 = b#(c(a(a(x1))))
    
    a(a(b(x1))) = 2x1 + 19 >= 2x1 + 13/2 = c(x1)
    
    a(c(x1)) = 4x1 + 14 >= 4x1 + 41/4 = b(c(a(a(x1))))
    
    b(c(x1)) = x1 + 29/4 >= x1 = x1
   problem:
    DPs:
     
    TRS:
     a(a(b(x1))) -> c(x1)
     a(c(x1)) -> b(c(a(a(x1))))
     b(c(x1)) -> x1
   Qed