0.58/0.62	MAYBE
0.58/0.62	
0.58/0.62	DP problem for innermost termination.
0.58/0.62	P =
0.58/0.62	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.58/0.62	  f2#(I0, I1) -> f2#(I0 + 1, I1 + 2) [I0 <= I1 /\ -1 <= I1 - 1 /\ I1 - I0 = 0 /\ -1 <= I0 - 1]
0.58/0.62	  f2#(I2, I3) -> f2#(I2 + 1, I3) [1 <= I3 - I2 /\ I2 <= I3 /\ -1 <= I2 - 1 /\ -1 <= I3 - 1]
0.58/0.62	  f1#(I4, I5) -> f2#(0, 0) 
0.58/0.62	R = 
0.58/0.62	  init(x1, x2) -> f1(rnd1, rnd2) 
0.58/0.62	  f2(I0, I1) -> f2(I0 + 1, I1 + 2) [I0 <= I1 /\ -1 <= I1 - 1 /\ I1 - I0 = 0 /\ -1 <= I0 - 1]
0.58/0.62	  f2(I2, I3) -> f2(I2 + 1, I3) [1 <= I3 - I2 /\ I2 <= I3 /\ -1 <= I2 - 1 /\ -1 <= I3 - 1]
0.58/0.62	  f1(I4, I5) -> f2(0, 0) 
0.58/0.62	
0.58/0.62	The dependency graph for this problem is:
0.58/0.62	  0 -> 3
0.58/0.62	  1 -> 2
0.58/0.62	  2 -> 1, 2
0.58/0.62	  3 -> 1
0.58/0.62	Where:
0.58/0.62	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.58/0.62	  1) f2#(I0, I1) -> f2#(I0 + 1, I1 + 2) [I0 <= I1 /\ -1 <= I1 - 1 /\ I1 - I0 = 0 /\ -1 <= I0 - 1]
0.58/0.62	  2) f2#(I2, I3) -> f2#(I2 + 1, I3) [1 <= I3 - I2 /\ I2 <= I3 /\ -1 <= I2 - 1 /\ -1 <= I3 - 1]
0.58/0.62	  3) f1#(I4, I5) -> f2#(0, 0) 
0.58/0.62	
0.58/0.62	We have the following SCCs.
0.58/0.62	  { 1, 2 }
0.58/0.62	
0.58/0.62	DP problem for innermost termination.
0.58/0.62	P =
0.58/0.62	  f2#(I0, I1) -> f2#(I0 + 1, I1 + 2) [I0 <= I1 /\ -1 <= I1 - 1 /\ I1 - I0 = 0 /\ -1 <= I0 - 1]
0.58/0.62	  f2#(I2, I3) -> f2#(I2 + 1, I3) [1 <= I3 - I2 /\ I2 <= I3 /\ -1 <= I2 - 1 /\ -1 <= I3 - 1]
0.58/0.62	R = 
0.58/0.62	  init(x1, x2) -> f1(rnd1, rnd2) 
0.58/0.62	  f2(I0, I1) -> f2(I0 + 1, I1 + 2) [I0 <= I1 /\ -1 <= I1 - 1 /\ I1 - I0 = 0 /\ -1 <= I0 - 1]
0.58/0.62	  f2(I2, I3) -> f2(I2 + 1, I3) [1 <= I3 - I2 /\ I2 <= I3 /\ -1 <= I2 - 1 /\ -1 <= I3 - 1]
0.58/0.62	  f1(I4, I5) -> f2(0, 0) 
0.58/0.62	
0.58/3.60	EOF