YES

Problem:
 a(a(b(x1))) -> c(b(a(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#(a(a(x1)))
   a#(c(x1)) -> a#(x1)
  TRS:
   a(a(b(x1))) -> c(b(a(a(a(x1)))))
   a(c(x1)) -> b(a(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
    [a#](x0) = [2 0]x0,
    
              [1 0]     [0]
    [c](x0) = [2 0]x0 + [2],
    
              [0  0 ]     [0]
    [a](x0) = [-& 0 ]x0 + [0],
    
              [-& -&]     [0]
    [b](x0) = [0  0 ]x0 + [2]
   orientation:
    a#(a(b(x1))) = [2 2]x1 + [4] >= [2 0]x1 = a#(x1)
    
    a#(a(b(x1))) = [2 2]x1 + [4] >= [2 2]x1 + [2] = a#(a(x1))
    
    a#(a(b(x1))) = [2 2]x1 + [4] >= [2 2]x1 + [2] = a#(a(a(x1)))
    
    a#(c(x1)) = [3 2]x1 + [2] >= [2 0]x1 = a#(x1)
    
                  [0 0]     [2]    [0 0]     [2]                    
    a(a(b(x1))) = [0 0]x1 + [2] >= [0 0]x1 + [2] = c(b(a(a(a(x1)))))
    
               [2 0]     [2]    [-& -&]     [0]           
    a(c(x1)) = [2 0]x1 + [2] >= [0  0 ]x1 + [2] = b(a(x1))
   problem:
    DPs:
     a#(a(b(x1))) -> a#(x1)
     a#(a(b(x1))) -> a#(a(x1))
     a#(a(b(x1))) -> a#(a(a(x1)))
    TRS:
     a(a(b(x1))) -> c(b(a(a(a(x1)))))
     a(c(x1)) -> b(a(x1))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
     [a#](x0) = [1 0 0]x0,
     
               [0 0 0]     [0]
     [c](x0) = [2 2 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 2]     [1]
    orientation:
     a#(a(b(x1))) = [1 1 2]x1 + [1] >= [1 0 0]x1 = a#(x1)
     
     a#(a(b(x1))) = [1 1 2]x1 + [1] >= [0 0 1]x1 = a#(a(x1))
     
     a#(a(b(x1))) = [1 1 2]x1 + [1] >= [0 1 0]x1 = a#(a(a(x1)))
     
                   [0 0 0]     [0]    [0 0 0]     [0]                    
     a(a(b(x1))) = [2 2 4]x1 + [2] >= [2 2 4]x1 + [2] = c(b(a(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 2 1]     [1]    [2 2 1]     [1]           
    problem:
     DPs:
      
     TRS:
      a(a(b(x1))) -> c(b(a(a(a(x1)))))
      a(c(x1)) -> b(a(x1))
    Qed