0.00/0.01	YES
0.00/0.01	
0.00/0.01	DP problem for innermost termination.
0.00/0.01	P =
0.00/0.01	  eval#(x, y) -> eval#(y, x) [x > y]
0.00/0.01	R = 
0.00/0.01	  eval(x, y) -> eval(y, x) [x > y]
0.00/0.01	
0.00/0.01	The dependency graph for this problem is:
0.00/0.01	  0 -> 
0.00/0.01	Where:
0.00/0.01	  0) eval#(x, y) -> eval#(y, x) [x > y]
0.00/0.01	
0.00/0.01	We have the following SCCs.
0.00/0.01	  
0.00/0.01	EOF