0.60/0.63	MAYBE
0.60/0.63	
0.60/0.63	DP problem for innermost termination.
0.60/0.63	P =
0.60/0.63	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.60/0.63	  f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1]
0.60/0.63	  f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2 /\ -1 <= I3 - 1]
0.60/0.63	  f2#(I6, I7) -> f2#(3, I8) [3 = I6]
0.60/0.63	  f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.60/0.63	R = 
0.60/0.63	  init(x1, x2) -> f1(rnd1, rnd2) 
0.60/0.63	  f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1]
0.60/0.63	  f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2 /\ -1 <= I3 - 1]
0.60/0.63	  f2(I6, I7) -> f2(3, I8) [3 = I6]
0.60/0.63	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.60/0.63	
0.60/0.63	The dependency graph for this problem is:
0.60/0.63	  0 -> 4
0.60/0.63	  1 -> 1
0.60/0.63	  2 -> 2, 3
0.60/0.63	  3 -> 3
0.60/0.63	  4 -> 1, 2, 3
0.60/0.63	Where:
0.60/0.63	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.60/0.63	  1) f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1]
0.60/0.63	  2) f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2 /\ -1 <= I3 - 1]
0.60/0.63	  3) f2#(I6, I7) -> f2#(3, I8) [3 = I6]
0.60/0.63	  4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.60/0.63	
0.60/0.63	We have the following SCCs.
0.60/0.63	  { 2 }
0.60/0.63	  { 3 }
0.60/0.63	  { 1 }
0.60/0.63	
0.60/0.63	DP problem for innermost termination.
0.60/0.63	P =
0.60/0.63	  f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1]
0.60/0.63	R = 
0.60/0.63	  init(x1, x2) -> f1(rnd1, rnd2) 
0.60/0.63	  f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1]
0.60/0.63	  f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2 /\ -1 <= I3 - 1]
0.60/0.63	  f2(I6, I7) -> f2(3, I8) [3 = I6]
0.60/0.63	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.60/0.63	
0.60/0.63	We use the reverse value criterion with the projection function NU:
0.60/0.63	  NU[f2#(z1,z2)] = 9 + -1 * z1
0.60/0.63	
0.60/0.63	This gives the following inequalities:
0.60/0.63	  I0 <= 9 /\ 3 <= I0 - 1  ==>  9 + -1 * I0 > 9 + -1 * (I0 + 1)  with  9 + -1 * I0 >= 0
0.60/0.63	
0.60/0.63	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.60/0.63	
0.60/0.63	DP problem for innermost termination.
0.60/0.63	P =
0.60/0.63	  f2#(I6, I7) -> f2#(3, I8) [3 = I6]
0.60/0.63	R = 
0.60/0.63	  init(x1, x2) -> f1(rnd1, rnd2) 
0.60/0.63	  f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1]
0.60/0.63	  f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2 /\ -1 <= I3 - 1]
0.60/0.63	  f2(I6, I7) -> f2(3, I8) [3 = I6]
0.60/0.63	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.60/0.63	
0.60/3.61	EOF