0.00/0.28	MAYBE
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  init#(x1) -> f1#(rnd1) 
0.00/0.28	  f2#(I0) -> f2#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.28	  f2#(I1) -> f2#(I1 - 1) [I1 - 1 <= I1 - 1 /\ I1 <= -1]
0.00/0.28	  f1#(I2) -> f2#(-1) 
0.00/0.28	R = 
0.00/0.28	  init(x1) -> f1(rnd1) 
0.00/0.28	  f2(I0) -> f2(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.28	  f2(I1) -> f2(I1 - 1) [I1 - 1 <= I1 - 1 /\ I1 <= -1]
0.00/0.28	  f1(I2) -> f2(-1) 
0.00/0.28	
0.00/0.28	The dependency graph for this problem is:
0.00/0.28	  0 -> 3
0.00/0.28	  1 -> 1
0.00/0.28	  2 -> 2
0.00/0.28	  3 -> 2
0.00/0.28	Where:
0.00/0.28	  0) init#(x1) -> f1#(rnd1) 
0.00/0.28	  1) f2#(I0) -> f2#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.28	  2) f2#(I1) -> f2#(I1 - 1) [I1 - 1 <= I1 - 1 /\ I1 <= -1]
0.00/0.28	  3) f1#(I2) -> f2#(-1) 
0.00/0.28	
0.00/0.28	We have the following SCCs.
0.00/0.28	  { 1 }
0.00/0.28	  { 2 }
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  f2#(I1) -> f2#(I1 - 1) [I1 - 1 <= I1 - 1 /\ I1 <= -1]
0.00/0.28	R = 
0.00/0.28	  init(x1) -> f1(rnd1) 
0.00/0.28	  f2(I0) -> f2(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.00/0.28	  f2(I1) -> f2(I1 - 1) [I1 - 1 <= I1 - 1 /\ I1 <= -1]
0.00/0.28	  f1(I2) -> f2(-1) 
0.00/0.28	
0.00/3.26	EOF