0.00/0.41	MAYBE
0.00/0.41	
0.00/0.41	DP problem for innermost termination.
0.00/0.41	P =
0.00/0.41	  init#(x1, x2) -> f2#(rnd1, rnd2) 
0.00/0.41	  f2#(I0, I1) -> f1#(17, 13) 
0.00/0.41	  f1#(I2, I3) -> f1#(I3, I2) [I2 + I3 <= 99 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.41	R = 
0.00/0.41	  init(x1, x2) -> f2(rnd1, rnd2) 
0.00/0.41	  f2(I0, I1) -> f1(17, 13) 
0.00/0.41	  f1(I2, I3) -> f1(I3, I2) [I2 + I3 <= 99 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.41	
0.00/0.41	The dependency graph for this problem is:
0.00/0.41	  0 -> 1
0.00/0.41	  1 -> 2
0.00/0.41	  2 -> 2
0.00/0.41	Where:
0.00/0.41	  0) init#(x1, x2) -> f2#(rnd1, rnd2) 
0.00/0.41	  1) f2#(I0, I1) -> f1#(17, 13) 
0.00/0.41	  2) f1#(I2, I3) -> f1#(I3, I2) [I2 + I3 <= 99 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.41	
0.00/0.41	We have the following SCCs.
0.00/0.41	  { 2 }
0.00/0.41	
0.00/0.41	DP problem for innermost termination.
0.00/0.41	P =
0.00/0.41	  f1#(I2, I3) -> f1#(I3, I2) [I2 + I3 <= 99 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.41	R = 
0.00/0.41	  init(x1, x2) -> f2(rnd1, rnd2) 
0.00/0.41	  f2(I0, I1) -> f1(17, 13) 
0.00/0.41	  f1(I2, I3) -> f1(I3, I2) [I2 + I3 <= 99 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.41	
0.00/3.39	EOF