YES

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

Proof:
 DP Processor:
  DPs:
   a#(a(b(x1))) -> a#(x1)
   a#(a(b(x1))) -> a#(a(x1))
   a#(a(b(x1))) -> a#(c(a(a(x1))))
   a#(c(x1)) -> a#(x1)
  TRS:
   a(a(b(x1))) -> c(a(c(a(a(x1)))))
   a(c(x1)) -> b(a(x1))
  EDG Processor:
   DPs:
    a#(a(b(x1))) -> a#(x1)
    a#(a(b(x1))) -> a#(a(x1))
    a#(a(b(x1))) -> a#(c(a(a(x1))))
    a#(c(x1)) -> a#(x1)
   TRS:
    a(a(b(x1))) -> c(a(c(a(a(x1)))))
    a(c(x1)) -> b(a(x1))
   graph:
    a#(c(x1)) -> a#(x1) -> a#(a(b(x1))) -> a#(x1)
    a#(c(x1)) -> a#(x1) -> a#(a(b(x1))) -> a#(a(x1))
    a#(c(x1)) -> a#(x1) -> a#(a(b(x1))) -> a#(c(a(a(x1))))
    a#(c(x1)) -> a#(x1) -> a#(c(x1)) -> a#(x1)
    a#(a(b(x1))) -> a#(c(a(a(x1)))) -> a#(c(x1)) -> a#(x1)
    a#(a(b(x1))) -> a#(a(x1)) -> a#(a(b(x1))) -> a#(x1)
    a#(a(b(x1))) -> a#(a(x1)) -> a#(a(b(x1))) -> a#(a(x1))
    a#(a(b(x1))) -> a#(a(x1)) -> a#(a(b(x1))) -> a#(c(a(a(x1))))
    a#(a(b(x1))) -> a#(a(x1)) -> a#(c(x1)) -> a#(x1)
    a#(a(b(x1))) -> a#(x1) -> a#(a(b(x1))) -> a#(x1)
    a#(a(b(x1))) -> a#(x1) -> a#(a(b(x1))) -> a#(a(x1))
    a#(a(b(x1))) -> a#(x1) -> a#(a(b(x1))) -> a#(c(a(a(x1))))
    a#(a(b(x1))) -> a#(x1) -> a#(c(x1)) -> a#(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
     [a#](x0) = [2 1 0]x0,
     
               [0 0 0]     [0]
     [c](x0) = [2 1 1]x0 + [1]
               [0 0 0]     [0],
     
               [0 0 1]  
     [a](x0) = [2 0 0]x0
               [0 1 0]  ,
     
               [0 0 0]     [0]
     [b](x0) = [0 0 0]x0 + [0]
               [1 1 1]     [1]
    orientation:
     a#(a(b(x1))) = [2 2 2]x1 + [2] >= [2 1 0]x1 = a#(x1)
     
     a#(a(b(x1))) = [2 2 2]x1 + [2] >= [2 0 2]x1 = a#(a(x1))
     
     a#(a(b(x1))) = [2 2 2]x1 + [2] >= [2 2 2]x1 + [1] = a#(c(a(a(x1))))
     
     a#(c(x1)) = [2 1 1]x1 + [1] >= [2 1 0]x1 = a#(x1)
     
                   [0 0 0]     [0]    [0 0 0]     [0]                    
     a(a(b(x1))) = [2 2 2]x1 + [2] >= [2 2 2]x1 + [2] = c(a(c(a(a(x1)))))
                   [0 0 0]     [0]    [0 0 0]     [0]                    
     
                [0 0 0]     [0]    [0 0 0]     [0]           
     a(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(x1))
                [2 1 1]     [1]    [2 1 1]     [1]           
    problem:
     DPs:
      
     TRS:
      a(a(b(x1))) -> c(a(c(a(a(x1)))))
      a(c(x1)) -> b(a(x1))
    Qed