YES

Problem:
 g(h(x1)) -> g(f(s(x1)))
 f(s(s(s(x1)))) -> h(f(s(h(x1))))
 f(h(x1)) -> h(f(s(h(x1))))
 h(x1) -> x1
 f(f(s(s(x1)))) -> s(s(s(f(f(x1)))))
 b(a(x1)) -> a(b(x1))
 a(a(a(x1))) -> b(a(a(b(x1))))
 b(b(b(b(x1)))) -> a(x1)

Proof:
 String Reversal Processor:
  h(g(x1)) -> s(f(g(x1)))
  s(s(s(f(x1)))) -> h(s(f(h(x1))))
  h(f(x1)) -> h(s(f(h(x1))))
  h(x1) -> x1
  s(s(f(f(x1)))) -> f(f(s(s(s(x1)))))
  a(b(x1)) -> b(a(x1))
  a(a(a(x1))) -> b(a(a(b(x1))))
  b(b(b(b(x1)))) -> a(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a](x0) = x0 + 2,
    
    [g](x0) = x0 + 8,
    
    [f](x0) = x0,
    
    [b](x0) = x0 + 1,
    
    [h](x0) = x0,
    
    [s](x0) = x0
   orientation:
    h(g(x1)) = x1 + 8 >= x1 + 8 = s(f(g(x1)))
    
    s(s(s(f(x1)))) = x1 >= x1 = h(s(f(h(x1))))
    
    h(f(x1)) = x1 >= x1 = h(s(f(h(x1))))
    
    h(x1) = x1 >= x1 = x1
    
    s(s(f(f(x1)))) = x1 >= x1 = f(f(s(s(s(x1)))))
    
    a(b(x1)) = x1 + 3 >= x1 + 3 = b(a(x1))
    
    a(a(a(x1))) = x1 + 6 >= x1 + 6 = b(a(a(b(x1))))
    
    b(b(b(b(x1)))) = x1 + 4 >= x1 + 2 = a(x1)
   problem:
    h(g(x1)) -> s(f(g(x1)))
    s(s(s(f(x1)))) -> h(s(f(h(x1))))
    h(f(x1)) -> h(s(f(h(x1))))
    h(x1) -> x1
    s(s(f(f(x1)))) -> f(f(s(s(s(x1)))))
    a(b(x1)) -> b(a(x1))
    a(a(a(x1))) -> b(a(a(b(x1))))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [a](x0) = x0 + 4,
     
     [g](x0) = 2x0 + 8,
     
     [f](x0) = x0,
     
     [b](x0) = x0,
     
     [h](x0) = x0,
     
     [s](x0) = x0
    orientation:
     h(g(x1)) = 2x1 + 8 >= 2x1 + 8 = s(f(g(x1)))
     
     s(s(s(f(x1)))) = x1 >= x1 = h(s(f(h(x1))))
     
     h(f(x1)) = x1 >= x1 = h(s(f(h(x1))))
     
     h(x1) = x1 >= x1 = x1
     
     s(s(f(f(x1)))) = x1 >= x1 = f(f(s(s(s(x1)))))
     
     a(b(x1)) = x1 + 4 >= x1 + 4 = b(a(x1))
     
     a(a(a(x1))) = x1 + 12 >= x1 + 8 = b(a(a(b(x1))))
    problem:
     h(g(x1)) -> s(f(g(x1)))
     s(s(s(f(x1)))) -> h(s(f(h(x1))))
     h(f(x1)) -> h(s(f(h(x1))))
     h(x1) -> x1
     s(s(f(f(x1)))) -> f(f(s(s(s(x1)))))
     a(b(x1)) -> b(a(x1))
    String Reversal Processor:
     g(h(x1)) -> g(f(s(x1)))
     f(s(s(s(x1)))) -> h(f(s(h(x1))))
     f(h(x1)) -> h(f(s(h(x1))))
     h(x1) -> x1
     f(f(s(s(x1)))) -> s(s(s(f(f(x1)))))
     b(a(x1)) -> a(b(x1))
     Bounds Processor:
      bound: 1
      enrichment: match
      automaton:
       final states: {14,9,2,5,1}
       transitions:
        g1(18) -> 19*
        s1(25) -> 26*
        s1(16) -> 17*
        g0(4) -> 1*
        f60() -> 2*
        b0(2) -> 15*
        a0(15) -> 14*
        f0(2) -> 10*
        f0(10) -> 11*
        f0(3) -> 4*
        f0(7) -> 8*
        h0(8) -> 5*
        h0(2) -> 6*
        s0(12) -> 13*
        s0(2) -> 3*
        s0(13) -> 9*
        s0(6) -> 7*
        s0(11) -> 12*
        h1(34) -> 35*
        h1(27) -> 28*
        h1(44) -> 45*
        h1(24) -> 25*
        h1(40) -> 41*
        f1(26) -> 27*
        f1(17) -> 18*
        27 -> 28,11
        19 -> 1*
        24 -> 25*
        35 -> 25*
        12 -> 40*
        14 -> 15*
        45 -> 25*
        28 -> 27,11
        11 -> 34*
        2 -> 6*
        44 -> 45,25
        13 -> 44*
        8 -> 24,16,5
        40 -> 41,25
        34 -> 35,25
        5 -> 10,4,8
        9 -> 10,11
        41 -> 25*
      problem:
       
      Qed