0.00/0.50	YES
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.50	  f3#(I0, I1) -> f2#(I2, I3) [I0 + I3 = I2 /\ 0 <= I1 - 2 * I3 /\ I1 - 2 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 <= I0 - 1 /\ I3 <= I1 - 1 /\ 0 <= I1 - 1]
0.00/0.50	  f2#(I4, I5) -> f3#(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	  f1#(I6, I7) -> f2#(0, I8) [0 <= I6 - 1 /\ 0 <= I7 - 1 /\ -1 <= I8 - 1]
0.00/0.50	R = 
0.00/0.50	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.50	  f3(I0, I1) -> f2(I2, I3) [I0 + I3 = I2 /\ 0 <= I1 - 2 * I3 /\ I1 - 2 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 <= I0 - 1 /\ I3 <= I1 - 1 /\ 0 <= I1 - 1]
0.00/0.50	  f2(I4, I5) -> f3(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	  f1(I6, I7) -> f2(0, I8) [0 <= I6 - 1 /\ 0 <= I7 - 1 /\ -1 <= I8 - 1]
0.00/0.50	
0.00/0.50	The dependency graph for this problem is:
0.00/0.50	  0 -> 3
0.00/0.50	  1 -> 2
0.00/0.50	  2 -> 1
0.00/0.50	  3 -> 2
0.00/0.50	Where:
0.00/0.50	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.50	  1) f3#(I0, I1) -> f2#(I2, I3) [I0 + I3 = I2 /\ 0 <= I1 - 2 * I3 /\ I1 - 2 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 <= I0 - 1 /\ I3 <= I1 - 1 /\ 0 <= I1 - 1]
0.00/0.50	  2) f2#(I4, I5) -> f3#(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	  3) f1#(I6, I7) -> f2#(0, I8) [0 <= I6 - 1 /\ 0 <= I7 - 1 /\ -1 <= I8 - 1]
0.00/0.50	
0.00/0.50	We have the following SCCs.
0.00/0.50	  { 1, 2 }
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  f3#(I0, I1) -> f2#(I2, I3) [I0 + I3 = I2 /\ 0 <= I1 - 2 * I3 /\ I1 - 2 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 <= I0 - 1 /\ I3 <= I1 - 1 /\ 0 <= I1 - 1]
0.00/0.50	  f2#(I4, I5) -> f3#(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	R = 
0.00/0.50	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.50	  f3(I0, I1) -> f2(I2, I3) [I0 + I3 = I2 /\ 0 <= I1 - 2 * I3 /\ I1 - 2 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 <= I0 - 1 /\ I3 <= I1 - 1 /\ 0 <= I1 - 1]
0.00/0.50	  f2(I4, I5) -> f3(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	  f1(I6, I7) -> f2(0, I8) [0 <= I6 - 1 /\ 0 <= I7 - 1 /\ -1 <= I8 - 1]
0.00/0.50	
0.00/0.50	We use the basic value criterion with the projection function NU:
0.00/0.50	  NU[f2#(z1,z2)] = z2
0.00/0.50	  NU[f3#(z1,z2)] = z2
0.00/0.50	
0.00/0.50	This gives the following inequalities:
0.00/0.50	  I0 + I3 = I2 /\ 0 <= I1 - 2 * I3 /\ I1 - 2 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 <= I0 - 1 /\ I3 <= I1 - 1 /\ 0 <= I1 - 1  ==>  I1 >! I3
0.00/0.50	  0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1  ==>  I5 (>! \union =) I5
0.00/0.50	
0.00/0.50	We remove all the strictly oriented dependency pairs.
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  f2#(I4, I5) -> f3#(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	R = 
0.00/0.50	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.50	  f3(I0, I1) -> f2(I2, I3) [I0 + I3 = I2 /\ 0 <= I1 - 2 * I3 /\ I1 - 2 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 <= I0 - 1 /\ I3 <= I1 - 1 /\ 0 <= I1 - 1]
0.00/0.50	  f2(I4, I5) -> f3(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	  f1(I6, I7) -> f2(0, I8) [0 <= I6 - 1 /\ 0 <= I7 - 1 /\ -1 <= I8 - 1]
0.00/0.50	
0.00/0.50	The dependency graph for this problem is:
0.00/0.50	  2 -> 
0.00/0.50	Where:
0.00/0.50	  2) f2#(I4, I5) -> f3#(I4, I5) [0 <= I5 - 1 /\ y1 <= I5 - 1 /\ -1 <= y1 - 1 /\ -1 <= I4 - 1]
0.00/0.50	
0.00/0.50	We have the following SCCs.
0.00/0.50	  
0.00/3.48	EOF