YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> a#(x1)
   a#(b(x1)) -> a#(a(x1))
   a#(b(x1)) -> a#(a(a(x1)))
   a#(b(x1)) -> b#(a(a(a(x1))))
   b#(a(x1)) -> a#(a(x1))
   a#(a(x1)) -> b#(x1)
   a#(a(x1)) -> a#(c(b(x1)))
  TRS:
   a(b(x1)) -> b(a(a(a(x1))))
   b(a(x1)) -> a(a(x1))
   a(a(x1)) -> a(c(b(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [b#](x0) = [2 0 1]x0,
    
    [a#](x0) = [1 1 1]x0,
    
              [0]
    [c](x0) = [0]
              [0],
    
              [0 1 0]     [1]
    [a](x0) = [0 1 0]x0 + [0]
              [2 0 1]     [0],
    
              [0 3 0]     [3]
    [b](x0) = [0 3 0]x0 + [3]
              [2 0 1]     [1]
   orientation:
    a#(b(x1)) = [2 6 1]x1 + [7] >= [1 1 1]x1 = a#(x1)
    
    a#(b(x1)) = [2 6 1]x1 + [7] >= [2 2 1]x1 + [1] = a#(a(x1))
    
    a#(b(x1)) = [2 6 1]x1 + [7] >= [2 4 1]x1 + [3] = a#(a(a(x1)))
    
    a#(b(x1)) = [2 6 1]x1 + [7] >= [2 6 1]x1 + [6] = b#(a(a(a(x1))))
    
    b#(a(x1)) = [2 2 1]x1 + [2] >= [2 2 1]x1 + [1] = a#(a(x1))
    
    a#(a(x1)) = [2 2 1]x1 + [1] >= [2 0 1]x1 = b#(x1)
    
    a#(a(x1)) = [2 2 1]x1 + [1] >= [0] = a#(c(b(x1)))
    
               [0 3 0]     [4]    [0 3 0]     [3]                 
    a(b(x1)) = [0 3 0]x1 + [3] >= [0 3 0]x1 + [3] = b(a(a(a(x1))))
               [2 6 1]     [7]    [2 6 1]     [7]                 
    
               [0 3 0]     [3]    [0 1 0]     [1]           
    b(a(x1)) = [0 3 0]x1 + [3] >= [0 1 0]x1 + [0] = a(a(x1))
               [2 2 1]     [3]    [2 2 1]     [2]           
    
               [0 1 0]     [1]    [1]              
    a(a(x1)) = [0 1 0]x1 + [0] >= [0] = a(c(b(x1)))
               [2 2 1]     [2]    [0]              
   problem:
    DPs:
     
    TRS:
     a(b(x1)) -> b(a(a(a(x1))))
     b(a(x1)) -> a(a(x1))
     a(a(x1)) -> a(c(b(x1)))
   Qed