YES

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

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