0.00/0.09	YES
0.00/0.09	
0.00/0.09	DP problem for innermost termination.
0.00/0.09	P =
0.00/0.09	  init#(x1) -> f1#(rnd1) 
0.00/0.09	  f2#(I0) -> f2#(I0 - 1) [0 <= I0 - 1]
0.00/0.09	  f1#(I1) -> f2#(9) 
0.00/0.09	R = 
0.00/0.09	  init(x1) -> f1(rnd1) 
0.00/0.09	  f2(I0) -> f2(I0 - 1) [0 <= I0 - 1]
0.00/0.09	  f1(I1) -> f2(9) 
0.00/0.09	
0.00/0.09	The dependency graph for this problem is:
0.00/0.09	  0 -> 2
0.00/0.09	  1 -> 1
0.00/0.09	  2 -> 1
0.00/0.09	Where:
0.00/0.09	  0) init#(x1) -> f1#(rnd1) 
0.00/0.09	  1) f2#(I0) -> f2#(I0 - 1) [0 <= I0 - 1]
0.00/0.09	  2) f1#(I1) -> f2#(9) 
0.00/0.09	
0.00/0.09	We have the following SCCs.
0.00/0.09	  { 1 }
0.00/0.09	
0.00/0.09	DP problem for innermost termination.
0.00/0.09	P =
0.00/0.09	  f2#(I0) -> f2#(I0 - 1) [0 <= I0 - 1]
0.00/0.09	R = 
0.00/0.09	  init(x1) -> f1(rnd1) 
0.00/0.09	  f2(I0) -> f2(I0 - 1) [0 <= I0 - 1]
0.00/0.09	  f1(I1) -> f2(9) 
0.00/0.09	
0.00/0.09	We use the basic value criterion with the projection function NU:
0.00/0.09	  NU[f2#(z1)] = z1
0.00/0.09	
0.00/0.09	This gives the following inequalities:
0.00/0.09	  0 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.09	
0.00/0.09	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.08	EOF