0.50/0.51	MAYBE
0.50/0.51	
0.50/0.51	DP problem for innermost termination.
0.50/0.51	P =
0.50/0.51	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.50/0.51	  f3#(I0, I1) -> f3#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.50/0.51	  f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1]
0.50/0.51	  f2#(I6, I7) -> f3#(I6, I8) [0 <= I6 - 1]
0.50/0.51	  f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.50/0.51	R = 
0.50/0.51	  init(x1, x2) -> f1(rnd1, rnd2) 
0.50/0.51	  f3(I0, I1) -> f3(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.50/0.51	  f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1]
0.50/0.51	  f2(I6, I7) -> f3(I6, I8) [0 <= I6 - 1]
0.50/0.51	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.50/0.51	
0.50/0.51	The dependency graph for this problem is:
0.50/0.51	  0 -> 4
0.50/0.51	  1 -> 1
0.50/0.51	  2 -> 2, 3
0.50/0.51	  3 -> 1
0.50/0.51	  4 -> 2, 3
0.50/0.51	Where:
0.50/0.51	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.50/0.51	  1) f3#(I0, I1) -> f3#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.50/0.51	  2) f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1]
0.50/0.51	  3) f2#(I6, I7) -> f3#(I6, I8) [0 <= I6 - 1]
0.50/0.51	  4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.50/0.51	
0.50/0.51	We have the following SCCs.
0.50/0.51	  { 2 }
0.50/0.51	  { 1 }
0.50/0.51	
0.50/0.51	DP problem for innermost termination.
0.50/0.51	P =
0.50/0.51	  f3#(I0, I1) -> f3#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.50/0.51	R = 
0.50/0.51	  init(x1, x2) -> f1(rnd1, rnd2) 
0.50/0.51	  f3(I0, I1) -> f3(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.50/0.51	  f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1]
0.50/0.51	  f2(I6, I7) -> f3(I6, I8) [0 <= I6 - 1]
0.50/0.51	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.50/0.51	
0.50/0.51	We use the basic value criterion with the projection function NU:
0.50/0.51	  NU[f3#(z1,z2)] = z1
0.50/0.51	
0.50/0.51	This gives the following inequalities:
0.50/0.51	  I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1  ==>  I0 >! I0 - 1
0.50/0.51	
0.50/0.51	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.50/0.51	
0.50/0.51	DP problem for innermost termination.
0.50/0.51	P =
0.50/0.51	  f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1]
0.50/0.51	R = 
0.50/0.51	  init(x1, x2) -> f1(rnd1, rnd2) 
0.50/0.51	  f3(I0, I1) -> f3(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1]
0.50/0.51	  f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1]
0.50/0.51	  f2(I6, I7) -> f3(I6, I8) [0 <= I6 - 1]
0.50/0.51	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.50/0.51	
0.50/0.51	EOF