0.76/0.79	MAYBE
0.76/0.79	
0.76/0.79	DP problem for innermost termination.
0.76/0.79	P =
0.76/0.79	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.76/0.79	  f2#(I0, I1) -> f2#(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I1 <= I0 - 1]
0.76/0.79	  f2#(I2, I3) -> f2#(I3, I3) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.76/0.79	  f2#(I4, I5) -> f2#(I4, I4 - 1) [I4 = I5 /\ 0 <= I4 - 1]
0.76/0.79	  f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ 1 <= I7 - 1 /\ -1 <= I8 - 1]
0.76/0.79	R = 
0.76/0.79	  init(x1, x2) -> f1(rnd1, rnd2) 
0.76/0.79	  f2(I0, I1) -> f2(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I1 <= I0 - 1]
0.76/0.79	  f2(I2, I3) -> f2(I3, I3) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.76/0.79	  f2(I4, I5) -> f2(I4, I4 - 1) [I4 = I5 /\ 0 <= I4 - 1]
0.76/0.79	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ 1 <= I7 - 1 /\ -1 <= I8 - 1]
0.76/0.79	
0.76/0.79	The dependency graph for this problem is:
0.76/0.79	  0 -> 4
0.76/0.79	  1 -> 2
0.76/0.79	  2 -> 3
0.76/0.79	  3 -> 1
0.76/0.79	  4 -> 1, 2, 3
0.76/0.79	Where:
0.76/0.79	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.76/0.79	  1) f2#(I0, I1) -> f2#(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I1 <= I0 - 1]
0.76/0.79	  2) f2#(I2, I3) -> f2#(I3, I3) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.76/0.79	  3) f2#(I4, I5) -> f2#(I4, I4 - 1) [I4 = I5 /\ 0 <= I4 - 1]
0.76/0.79	  4) f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ 1 <= I7 - 1 /\ -1 <= I8 - 1]
0.76/0.79	
0.76/0.79	We have the following SCCs.
0.76/0.79	  { 1, 2, 3 }
0.76/0.79	
0.76/0.79	DP problem for innermost termination.
0.76/0.79	P =
0.76/0.79	  f2#(I0, I1) -> f2#(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I1 <= I0 - 1]
0.76/0.79	  f2#(I2, I3) -> f2#(I3, I3) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.76/0.79	  f2#(I4, I5) -> f2#(I4, I4 - 1) [I4 = I5 /\ 0 <= I4 - 1]
0.76/0.79	R = 
0.76/0.79	  init(x1, x2) -> f1(rnd1, rnd2) 
0.76/0.79	  f2(I0, I1) -> f2(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I1 <= I0 - 1]
0.76/0.79	  f2(I2, I3) -> f2(I3, I3) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.76/0.79	  f2(I4, I5) -> f2(I4, I4 - 1) [I4 = I5 /\ 0 <= I4 - 1]
0.76/0.79	  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I9 - 1 /\ 1 <= I7 - 1 /\ -1 <= I8 - 1]
0.76/0.79	
0.76/3.77	EOF