0.00/0.26	MAYBE
0.00/0.26	
0.00/0.26	DP problem for innermost termination.
0.00/0.26	P =
0.00/0.26	  init#(x1) -> f1#(rnd1) 
0.00/0.26	  f2#(I0) -> f2#(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.26	  f1#(I2) -> f2#(I3) [-1 <= I3 - 1]
0.00/0.26	R = 
0.00/0.26	  init(x1) -> f1(rnd1) 
0.00/0.26	  f2(I0) -> f2(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.26	  f1(I2) -> f2(I3) [-1 <= I3 - 1]
0.00/0.26	
0.00/0.26	The dependency graph for this problem is:
0.00/0.26	  0 -> 2
0.00/0.26	  1 -> 1
0.00/0.26	  2 -> 1
0.00/0.26	Where:
0.00/0.26	  0) init#(x1) -> f1#(rnd1) 
0.00/0.26	  1) f2#(I0) -> f2#(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.26	  2) f1#(I2) -> f2#(I3) [-1 <= I3 - 1]
0.00/0.26	
0.00/0.26	We have the following SCCs.
0.00/0.26	  { 1 }
0.00/0.26	
0.00/0.26	DP problem for innermost termination.
0.00/0.26	P =
0.00/0.26	  f2#(I0) -> f2#(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.26	R = 
0.00/0.26	  init(x1) -> f1(rnd1) 
0.00/0.26	  f2(I0) -> f2(I1) [-1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.26	  f1(I2) -> f2(I3) [-1 <= I3 - 1]
0.00/0.26	
0.00/3.24	EOF