YES

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

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