0.00/0.47	YES
0.00/0.47	
0.00/0.47	DP problem for innermost termination.
0.00/0.47	P =
0.00/0.47	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.47	  f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
0.00/0.47	  f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	  f1#(I5, I6) -> f2#(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]
0.00/0.47	R = 
0.00/0.47	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.47	  f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
0.00/0.47	  f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	  f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]
0.00/0.47	
0.00/0.47	The dependency graph for this problem is:
0.00/0.47	  0 -> 3
0.00/0.47	  1 -> 2
0.00/0.47	  2 -> 1
0.00/0.47	  3 -> 2
0.00/0.47	Where:
0.00/0.47	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.47	  1) f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
0.00/0.47	  2) f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	  3) f1#(I5, I6) -> f2#(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]
0.00/0.47	
0.00/0.47	We have the following SCCs.
0.00/0.47	  { 1, 2 }
0.00/0.47	
0.00/0.47	DP problem for innermost termination.
0.00/0.47	P =
0.00/0.47	  f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
0.00/0.47	  f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	R = 
0.00/0.47	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.47	  f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
0.00/0.47	  f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	  f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]
0.00/0.47	
0.00/0.47	We use the basic value criterion with the projection function NU:
0.00/0.47	  NU[f2#(z1,z2)] = z2
0.00/0.47	  NU[f3#(z1,z2)] = z2
0.00/0.47	
0.00/0.47	This gives the following inequalities:
0.00/0.47	  0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2  ==>  I1 >! I2
0.00/0.47	  0 <= I3 - 1 /\ 0 <= I4 - 1  ==>  I4 (>! \union =) I4
0.00/0.47	
0.00/0.47	We remove all the strictly oriented dependency pairs.
0.00/0.47	
0.00/0.47	DP problem for innermost termination.
0.00/0.47	P =
0.00/0.47	  f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	R = 
0.00/0.47	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.47	  f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2]
0.00/0.47	  f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	  f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1]
0.00/0.47	
0.00/0.47	The dependency graph for this problem is:
0.00/0.47	  2 -> 
0.00/0.47	Where:
0.00/0.47	  2) f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1]
0.00/0.47	
0.00/0.47	We have the following SCCs.
0.00/0.47	  
0.00/3.45	EOF