YES

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

Proof:
 DP Processor:
  DPs:
   a#(a(d(d(x1)))) -> b#(x1)
   a#(a(d(d(x1)))) -> b#(b(x1))
   a#(a(x1)) -> b#(x1)
   a#(a(x1)) -> b#(b(x1))
   a#(a(x1)) -> b#(b(b(x1)))
   a#(a(x1)) -> b#(b(b(b(x1))))
   a#(a(x1)) -> b#(b(b(b(b(x1)))))
   a#(a(x1)) -> b#(b(b(b(b(b(x1))))))
   b#(b(d(d(b(b(x1)))))) -> c#(x1)
   b#(b(d(d(b(b(x1)))))) -> c#(c(x1))
   b#(b(d(d(b(b(x1)))))) -> a#(c(c(x1)))
   b#(b(d(d(b(b(x1)))))) -> a#(a(c(c(x1))))
  TRS:
   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c#](x0) = 5x0,
    
    [b#](x0) = 4x0 + 5,
    
    [a#](x0) = 4x0 + 24,
    
    [c](x0) = 2x0 + 1,
    
    [b](x0) = x0 + 1,
    
    [a](x0) = x0 + 4,
    
    [d](x0) = 2x0 + 1
   orientation:
    a#(a(d(d(x1)))) = 16x1 + 52 >= 4x1 + 5 = b#(x1)
    
    a#(a(d(d(x1)))) = 16x1 + 52 >= 4x1 + 9 = b#(b(x1))
    
    a#(a(x1)) = 4x1 + 40 >= 4x1 + 5 = b#(x1)
    
    a#(a(x1)) = 4x1 + 40 >= 4x1 + 9 = b#(b(x1))
    
    a#(a(x1)) = 4x1 + 40 >= 4x1 + 13 = b#(b(b(x1)))
    
    a#(a(x1)) = 4x1 + 40 >= 4x1 + 17 = b#(b(b(b(x1))))
    
    a#(a(x1)) = 4x1 + 40 >= 4x1 + 21 = b#(b(b(b(b(x1)))))
    
    a#(a(x1)) = 4x1 + 40 >= 4x1 + 25 = b#(b(b(b(b(b(x1))))))
    
    b#(b(d(d(b(b(x1)))))) = 16x1 + 53 >= 5x1 = c#(x1)
    
    b#(b(d(d(b(b(x1)))))) = 16x1 + 53 >= 10x1 + 5 = c#(c(x1))
    
    b#(b(d(d(b(b(x1)))))) = 16x1 + 53 >= 16x1 + 36 = a#(c(c(x1)))
    
    b#(b(d(d(b(b(x1)))))) = 16x1 + 53 >= 16x1 + 52 = a#(a(c(c(x1))))
    
    a(a(d(d(x1)))) = 4x1 + 11 >= 4x1 + 11 = d(d(b(b(x1))))
    
    a(a(x1)) = x1 + 8 >= x1 + 6 = b(b(b(b(b(b(x1))))))
    
    b(b(d(d(b(b(x1)))))) = 4x1 + 13 >= 4x1 + 11 = a(a(c(c(x1))))
    
    c(c(x1)) = 4x1 + 3 >= 4x1 + 3 = d(d(x1))
   problem:
    DPs:
     
    TRS:
     a(a(d(d(x1)))) -> d(d(b(b(x1))))
     a(a(x1)) -> b(b(b(b(b(b(x1))))))
     b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
     c(c(x1)) -> d(d(x1))
   Qed