0.00/0.18	YES
0.00/0.18	
0.00/0.18	DP problem for innermost termination.
0.00/0.18	P =
0.00/0.18	  init#(x1) -> f1#(rnd1) 
0.00/0.18	  f2#(I0) -> f2#(I0 + 1) [I0 <= 10]
0.00/0.18	  f1#(I1) -> f2#(0) 
0.00/0.18	R = 
0.00/0.18	  init(x1) -> f1(rnd1) 
0.00/0.18	  f2(I0) -> f2(I0 + 1) [I0 <= 10]
0.00/0.18	  f1(I1) -> f2(0) 
0.00/0.18	
0.00/0.18	The dependency graph for this problem is:
0.00/0.18	  0 -> 2
0.00/0.18	  1 -> 1
0.00/0.18	  2 -> 1
0.00/0.18	Where:
0.00/0.18	  0) init#(x1) -> f1#(rnd1) 
0.00/0.18	  1) f2#(I0) -> f2#(I0 + 1) [I0 <= 10]
0.00/0.18	  2) f1#(I1) -> f2#(0) 
0.00/0.18	
0.00/0.18	We have the following SCCs.
0.00/0.18	  { 1 }
0.00/0.18	
0.00/0.18	DP problem for innermost termination.
0.00/0.18	P =
0.00/0.18	  f2#(I0) -> f2#(I0 + 1) [I0 <= 10]
0.00/0.18	R = 
0.00/0.18	  init(x1) -> f1(rnd1) 
0.00/0.18	  f2(I0) -> f2(I0 + 1) [I0 <= 10]
0.00/0.18	  f1(I1) -> f2(0) 
0.00/0.18	
0.00/0.18	We use the reverse value criterion with the projection function NU:
0.00/0.18	  NU[f2#(z1)] = 10 + -1 * z1
0.00/0.18	
0.00/0.18	This gives the following inequalities:
0.00/0.18	  I0 <= 10  ==>  10 + -1 * I0 > 10 + -1 * (I0 + 1)  with  10 + -1 * I0 >= 0
0.00/0.18	
0.00/0.18	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.16	EOF