0.00/0.45	MAYBE
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.45	  f2#(I0, I1) -> f2#(I0 - 1, I2) [5 <= I0 - 1 /\ I0 <= 9]
0.00/0.45	  f2#(I3, I4) -> f2#(I3, I5) [9 <= I3 - 1]
0.00/0.45	  f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I0 - 1, I2) [5 <= I0 - 1 /\ I0 <= 9]
0.00/0.45	  f2(I3, I4) -> f2(I3, I5) [9 <= I3 - 1]
0.00/0.45	  f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.45	
0.00/0.45	The dependency graph for this problem is:
0.00/0.45	  0 -> 3
0.00/0.45	  1 -> 1
0.00/0.45	  2 -> 2
0.00/0.45	  3 -> 1, 2
0.00/0.45	Where:
0.00/0.45	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.45	  1) f2#(I0, I1) -> f2#(I0 - 1, I2) [5 <= I0 - 1 /\ I0 <= 9]
0.00/0.45	  2) f2#(I3, I4) -> f2#(I3, I5) [9 <= I3 - 1]
0.00/0.45	  3) f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.45	
0.00/0.45	We have the following SCCs.
0.00/0.45	  { 2 }
0.00/0.45	  { 1 }
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  f2#(I0, I1) -> f2#(I0 - 1, I2) [5 <= I0 - 1 /\ I0 <= 9]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I0 - 1, I2) [5 <= I0 - 1 /\ I0 <= 9]
0.00/0.45	  f2(I3, I4) -> f2(I3, I5) [9 <= I3 - 1]
0.00/0.45	  f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.45	
0.00/0.45	We use the basic value criterion with the projection function NU:
0.00/0.45	  NU[f2#(z1,z2)] = z1
0.00/0.45	
0.00/0.45	This gives the following inequalities:
0.00/0.45	  5 <= I0 - 1 /\ I0 <= 9  ==>  I0 >! I0 - 1
0.00/0.45	
0.00/0.45	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  f2#(I3, I4) -> f2#(I3, I5) [9 <= I3 - 1]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I0 - 1, I2) [5 <= I0 - 1 /\ I0 <= 9]
0.00/0.45	  f2(I3, I4) -> f2(I3, I5) [9 <= I3 - 1]
0.00/0.45	  f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.45	
0.45/3.43	EOF