1.30/0.64	YES
1.30/0.65	
1.30/0.65	Problem:
1.30/0.65	 b(a(b(a(c(b(a(x1))))))) -> a(b(a(c(b(b(a(b(a(c(x1))))))))))
1.30/0.65	
1.30/0.65	Proof:
1.30/0.65	 String Reversal Processor:
1.30/0.65	  a(b(c(a(b(a(b(x1))))))) -> c(a(b(a(b(b(c(a(b(a(x1))))))))))
1.30/0.65	  Bounds Processor:
1.30/0.65	   bound: 1
1.30/0.65	   enrichment: match
1.30/0.65	   automaton:
1.30/0.65	    final states: {1}
1.30/0.65	    transitions:
1.30/0.65	     f30() -> 2*
1.30/0.65	     c0(5) -> 6*
1.30/0.65	     c0(11) -> 1*
1.30/0.65	     a0(10) -> 11*
1.30/0.65	     a0(2) -> 3*
1.30/0.65	     a0(4) -> 5*
1.30/0.65	     a0(8) -> 9*
1.30/0.65	     b0(7) -> 8*
1.30/0.65	     b0(9) -> 10*
1.30/0.65	     b0(6) -> 7*
1.30/0.65	     b0(3) -> 4*
1.30/0.65	     c1(15) -> 16*
1.30/0.65	     c1(21) -> 22*
1.30/0.65	     a1(20) -> 21*
1.30/0.65	     a1(12) -> 13*
1.30/0.65	     a1(14) -> 15*
1.30/0.65	     a1(18) -> 19*
1.30/0.65	     b1(17) -> 18*
1.30/0.65	     b1(19) -> 20*
1.30/0.65	     b1(16) -> 17*
1.30/0.65	     b1(13) -> 14*
1.30/0.65	     1 -> 13,3
1.30/0.65	     7 -> 12*
1.30/0.65	     22 -> 15,5
1.30/0.65	   problem:
1.30/0.65	    
1.30/0.65	   Qed
1.30/0.65	EOF