1.90/0.80	YES
1.90/0.80	
1.90/0.80	Problem:
1.90/0.80	 a(b(x1)) -> b(b(b(x1)))
1.90/0.80	 b(a(x1)) -> a(a(a(x1)))
1.90/0.80	 a(x1) -> x1
1.90/0.80	 b(x1) -> x1
1.90/0.80	
1.90/0.80	Proof:
1.90/0.80	 String Reversal Processor:
1.90/0.80	  b(a(x1)) -> b(b(b(x1)))
1.90/0.80	  a(b(x1)) -> a(a(a(x1)))
1.90/0.80	  a(x1) -> x1
1.90/0.80	  b(x1) -> x1
1.90/0.80	  Bounds Processor:
1.90/0.80	   bound: 0
1.90/0.80	   enrichment: match
1.90/0.80	   automaton:
1.90/0.80	    final states: {2,5,1}
1.90/0.80	    transitions:
1.90/0.80	     f20() -> 2*
1.90/0.80	     b0(2) -> 3*
1.90/0.80	     b0(4) -> 1*
1.90/0.80	     b0(3) -> 4*
1.90/0.80	     a0(7) -> 5*
1.90/0.80	     a0(2) -> 6*
1.90/0.80	     a0(6) -> 7*
1.90/0.80	     1 -> 3,4
1.90/0.80	     2 -> 3,6
1.90/0.80	     3 -> 4*
1.90/0.80	     4 -> 1*
1.90/0.80	     5 -> 6,7
1.90/0.80	     6 -> 7*
1.90/0.80	     7 -> 5*
1.90/0.80	   problem:
1.90/0.80	    
1.90/0.80	   Qed
1.90/0.80	EOF