0.38/0.76	MAYBE
0.38/0.76	
0.38/0.76	DP problem for innermost termination.
0.38/0.76	P =
0.38/0.76	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.38/0.76	  f2#(I0, I1, I2) -> f2#(I0 + 1, I1, I0 + 2) [-2 <= I0 - 1 /\ I2 <= I1]
0.38/0.76	  f1#(I3, I4, I5) -> f2#(I6, I7, I8) [I6 + 1 = I8 /\ 0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.38/0.76	R = 
0.38/0.76	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.38/0.76	  f2(I0, I1, I2) -> f2(I0 + 1, I1, I0 + 2) [-2 <= I0 - 1 /\ I2 <= I1]
0.38/0.76	  f1(I3, I4, I5) -> f2(I6, I7, I8) [I6 + 1 = I8 /\ 0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.38/0.76	
0.38/0.76	The dependency graph for this problem is:
0.38/0.76	  0 -> 2
0.38/0.76	  1 -> 1
0.38/0.76	  2 -> 1
0.38/0.76	Where:
0.38/0.76	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.38/0.76	  1) f2#(I0, I1, I2) -> f2#(I0 + 1, I1, I0 + 2) [-2 <= I0 - 1 /\ I2 <= I1]
0.38/0.76	  2) f1#(I3, I4, I5) -> f2#(I6, I7, I8) [I6 + 1 = I8 /\ 0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.38/0.76	
0.38/0.76	We have the following SCCs.
0.38/0.76	  { 1 }
0.38/0.76	
0.38/0.76	DP problem for innermost termination.
0.38/0.76	P =
0.38/0.76	  f2#(I0, I1, I2) -> f2#(I0 + 1, I1, I0 + 2) [-2 <= I0 - 1 /\ I2 <= I1]
0.38/0.76	R = 
0.38/0.76	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.38/0.76	  f2(I0, I1, I2) -> f2(I0 + 1, I1, I0 + 2) [-2 <= I0 - 1 /\ I2 <= I1]
0.38/0.76	  f1(I3, I4, I5) -> f2(I6, I7, I8) [I6 + 1 = I8 /\ 0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.38/0.76	
0.38/3.74	EOF