0.00/0.33	MAYBE
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  init#(x1, x2) -> f2#(rnd1, rnd2) 
0.00/0.33	  f2#(I0, I1) -> f1#(I2, I1) [0 <= I2 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I2 <= I0]
0.00/0.33	  f1#(I3, I4) -> f1#(I5, I4) [0 <= I5 - 1 /\ 0 <= I3 - 1 /\ I5 <= I3]
0.00/0.33	R = 
0.00/0.33	  init(x1, x2) -> f2(rnd1, rnd2) 
0.00/0.33	  f2(I0, I1) -> f1(I2, I1) [0 <= I2 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I2 <= I0]
0.00/0.33	  f1(I3, I4) -> f1(I5, I4) [0 <= I5 - 1 /\ 0 <= I3 - 1 /\ I5 <= I3]
0.00/0.33	
0.00/0.33	The dependency graph for this problem is:
0.00/0.33	  0 -> 1
0.00/0.33	  1 -> 2
0.00/0.33	  2 -> 2
0.00/0.33	Where:
0.00/0.33	  0) init#(x1, x2) -> f2#(rnd1, rnd2) 
0.00/0.33	  1) f2#(I0, I1) -> f1#(I2, I1) [0 <= I2 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I2 <= I0]
0.00/0.33	  2) f1#(I3, I4) -> f1#(I5, I4) [0 <= I5 - 1 /\ 0 <= I3 - 1 /\ I5 <= I3]
0.00/0.33	
0.00/0.33	We have the following SCCs.
0.00/0.33	  { 2 }
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  f1#(I3, I4) -> f1#(I5, I4) [0 <= I5 - 1 /\ 0 <= I3 - 1 /\ I5 <= I3]
0.00/0.33	R = 
0.00/0.33	  init(x1, x2) -> f2(rnd1, rnd2) 
0.00/0.33	  f2(I0, I1) -> f1(I2, I1) [0 <= I2 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I2 <= I0]
0.00/0.33	  f1(I3, I4) -> f1(I5, I4) [0 <= I5 - 1 /\ 0 <= I3 - 1 /\ I5 <= I3]
0.00/0.33	
0.00/3.31	EOF