0.00/0.38	YES
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 2147483647]
0.00/0.38	  f1#(I3, I4) -> f2#(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 2147483647]
0.00/0.38	  f1(I3, I4) -> f2(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  0 -> 2
0.00/0.38	  1 -> 1
0.00/0.38	  2 -> 1
0.00/0.38	Where:
0.00/0.38	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  1) f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 2147483647]
0.00/0.38	  2) f1#(I3, I4) -> f2#(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  { 1 }
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 2147483647]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 2147483647]
0.00/0.38	  f1(I3, I4) -> f2(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1]
0.00/0.38	
0.00/0.38	We use the reverse value criterion with the projection function NU:
0.00/0.38	  NU[f2#(z1,z2)] = 2147483647 + -1 * z1
0.00/0.38	
0.00/0.38	This gives the following inequalities:
0.00/0.38	  I0 <= 2147483647  ==>  2147483647 + -1 * I0 > 2147483647 + -1 * (I0 + 1)  with  2147483647 + -1 * I0 >= 0
0.00/0.38	
0.00/0.38	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.36	EOF