0.00/0.07	YES
0.00/0.07	
0.00/0.07	DP problem for innermost termination.
0.00/0.07	P =
0.00/0.07	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.07	  f2#(I0, I1) -> f2#(I1, I0) [I1 <= I0 - 1]
0.00/0.07	  f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1]
0.00/0.07	R = 
0.00/0.07	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.07	  f2(I0, I1) -> f2(I1, I0) [I1 <= I0 - 1]
0.00/0.07	  f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1]
0.00/0.07	
0.00/0.07	The dependency graph for this problem is:
0.00/0.07	  0 -> 2
0.00/0.07	  1 -> 
0.00/0.07	  2 -> 1
0.00/0.07	Where:
0.00/0.07	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.07	  1) f2#(I0, I1) -> f2#(I1, I0) [I1 <= I0 - 1]
0.00/0.07	  2) f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1]
0.00/0.07	
0.00/0.07	We have the following SCCs.
0.00/0.07	  
0.00/3.05	EOF