YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
  TRS:
   a(b(a(x1))) -> b(b(b(b(x1))))
   a(b(b(x1))) -> b(b(a(a(x1))))
  Matrix Interpretation Processor: dim=5
   
   interpretation:
    [a#](x0) = [0 1 0 0 0]x0,
    
              [1 0 0 0 0]     [0]
              [0 0 0 0 1]     [1]
    [b](x0) = [0 1 0 0 0]x0 + [0]
              [1 0 1 0 0]     [0]
              [0 1 0 0 0]     [0],
    
              [1 0 0 0 0]     [0]
              [0 1 0 0 0]     [0]
    [a](x0) = [0 1 0 1 1]x0 + [0]
              [1 1 0 0 0]     [0]
              [0 1 0 1 0]     [1]
   orientation:
    a#(b(b(x1))) = [0 1 0 0 0]x1 + [1] >= [0 1 0 0 0]x1 = a#(x1)
    
    a#(b(b(x1))) = [0 1 0 0 0]x1 + [1] >= [0 1 0 0 0]x1 = a#(a(x1))
    
                  [1 0 0 0 0]     [0]    [1 0 0 0 0]     [0]                 
                  [0 1 0 1 0]     [2]    [0 1 0 0 0]     [2]                 
    a(b(a(x1))) = [1 3 0 2 1]x1 + [2] >= [0 0 0 0 1]x1 + [2] = b(b(b(b(x1))))
                  [1 1 0 1 0]     [2]    [1 1 0 0 0]     [1]                 
                  [1 2 0 2 1]     [3]    [0 0 0 0 1]     [2]                 
    
                  [1 0 0 0 0]     [0]    [1 0 0 0 0]     [0]                 
                  [0 1 0 0 0]     [1]    [0 1 0 0 0]     [1]                 
    a(b(b(x1))) = [1 2 0 0 1]x1 + [2] >= [1 2 0 0 0]x1 + [2] = b(b(a(a(x1))))
                  [1 1 0 0 0]     [1]    [1 1 0 0 0]     [0]                 
                  [1 2 0 0 0]     [2]    [1 2 0 0 0]     [2]                 
   problem:
    DPs:
     
    TRS:
     a(b(a(x1))) -> b(b(b(b(x1))))
     a(b(b(x1))) -> b(b(a(a(x1))))
   Qed