0.00/0.34	MAYBE
0.00/0.34	
0.00/0.34	DP problem for innermost termination.
0.00/0.34	P =
0.00/0.34	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.34	  f2#(I0, I1) -> f2#(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.34	  f1#(I2, I3) -> f2#(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]
0.00/0.34	R = 
0.00/0.34	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.34	  f2(I0, I1) -> f2(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.34	  f1(I2, I3) -> f2(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]
0.00/0.34	
0.00/0.34	The dependency graph for this problem is:
0.00/0.34	  0 -> 2
0.00/0.34	  1 -> 1
0.00/0.34	  2 -> 1
0.00/0.34	Where:
0.00/0.34	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.34	  1) f2#(I0, I1) -> f2#(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.34	  2) f1#(I2, I3) -> f2#(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]
0.00/0.34	
0.00/0.34	We have the following SCCs.
0.00/0.34	  { 1 }
0.00/0.34	
0.00/0.34	DP problem for innermost termination.
0.00/0.34	P =
0.00/0.34	  f2#(I0, I1) -> f2#(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.34	R = 
0.00/0.34	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.34	  f2(I0, I1) -> f2(I1, I0) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.34	  f1(I2, I3) -> f2(I3 + 5, I3) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I3 + 5 - 1]
0.00/0.34	
0.00/3.33	EOF