0.00/0.55	MAYBE
0.00/0.55	
0.00/0.55	DP problem for innermost termination.
0.00/0.55	P =
0.00/0.55	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.55	  f2#(I0, I1) -> f2#(1, I2) [0 = I0]
0.00/0.55	  f2#(I3, I4) -> f2#(0, I5) [0 <= I3 - 1]
0.00/0.55	  f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.55	R = 
0.00/0.55	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.55	  f2(I0, I1) -> f2(1, I2) [0 = I0]
0.00/0.55	  f2(I3, I4) -> f2(0, I5) [0 <= I3 - 1]
0.00/0.55	  f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.55	
0.00/0.55	The dependency graph for this problem is:
0.00/0.55	  0 -> 3
0.00/0.55	  1 -> 2
0.00/0.55	  2 -> 1
0.00/0.55	  3 -> 1, 2
0.00/0.55	Where:
0.00/0.55	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.55	  1) f2#(I0, I1) -> f2#(1, I2) [0 = I0]
0.00/0.55	  2) f2#(I3, I4) -> f2#(0, I5) [0 <= I3 - 1]
0.00/0.55	  3) f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.55	
0.00/0.55	We have the following SCCs.
0.00/0.55	  { 1, 2 }
0.00/0.55	
0.00/0.55	DP problem for innermost termination.
0.00/0.55	P =
0.00/0.55	  f2#(I0, I1) -> f2#(1, I2) [0 = I0]
0.00/0.55	  f2#(I3, I4) -> f2#(0, I5) [0 <= I3 - 1]
0.00/0.55	R = 
0.00/0.55	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.55	  f2(I0, I1) -> f2(1, I2) [0 = I0]
0.00/0.55	  f2(I3, I4) -> f2(0, I5) [0 <= I3 - 1]
0.00/0.55	  f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.55	
0.51/3.52	EOF