0.00/0.42	MAYBE
0.00/0.42	
0.00/0.42	DP problem for innermost termination.
0.00/0.42	P =
0.00/0.42	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.42	  f3#(I0, I1) -> f3#(I0 - 1, I1 + I0) [0 <= I1 - 1]
0.00/0.42	  f2#(I2, I3) -> f2#(I2 - 1, I3 + I2) [0 <= I3 - 1]
0.00/0.42	  f1#(I4, I5) -> f3#(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ 1 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.42	  f1#(I8, I9) -> f2#(0, I10) [1 = I9 /\ -1 <= I10 - 1 /\ 0 <= I8 - 1]
0.00/0.42	R = 
0.00/0.42	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.42	  f3(I0, I1) -> f3(I0 - 1, I1 + I0) [0 <= I1 - 1]
0.00/0.42	  f2(I2, I3) -> f2(I2 - 1, I3 + I2) [0 <= I3 - 1]
0.00/0.42	  f1(I4, I5) -> f3(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ 1 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.42	  f1(I8, I9) -> f2(0, I10) [1 = I9 /\ -1 <= I10 - 1 /\ 0 <= I8 - 1]
0.00/0.42	
0.00/0.42	The dependency graph for this problem is:
0.00/0.42	  0 -> 3, 4
0.00/0.42	  1 -> 1
0.00/0.42	  2 -> 2
0.00/0.42	  3 -> 1
0.00/0.42	  4 -> 2
0.00/0.42	Where:
0.00/0.42	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.42	  1) f3#(I0, I1) -> f3#(I0 - 1, I1 + I0) [0 <= I1 - 1]
0.00/0.42	  2) f2#(I2, I3) -> f2#(I2 - 1, I3 + I2) [0 <= I3 - 1]
0.00/0.42	  3) f1#(I4, I5) -> f3#(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ 1 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.42	  4) f1#(I8, I9) -> f2#(0, I10) [1 = I9 /\ -1 <= I10 - 1 /\ 0 <= I8 - 1]
0.00/0.42	
0.00/0.42	We have the following SCCs.
0.00/0.42	  { 1 }
0.00/0.42	  { 2 }
0.00/0.42	
0.00/0.42	DP problem for innermost termination.
0.00/0.42	P =
0.00/0.42	  f2#(I2, I3) -> f2#(I2 - 1, I3 + I2) [0 <= I3 - 1]
0.00/0.42	R = 
0.00/0.42	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.42	  f3(I0, I1) -> f3(I0 - 1, I1 + I0) [0 <= I1 - 1]
0.00/0.42	  f2(I2, I3) -> f2(I2 - 1, I3 + I2) [0 <= I3 - 1]
0.00/0.42	  f1(I4, I5) -> f3(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ 1 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.42	  f1(I8, I9) -> f2(0, I10) [1 = I9 /\ -1 <= I10 - 1 /\ 0 <= I8 - 1]
0.00/0.42	
0.00/3.39	EOF