0.00/0.52	YES
0.00/0.52	
0.00/0.52	DP problem for innermost termination.
0.00/0.52	P =
0.00/0.52	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.52	  f3#(I0, I1) -> f2#(I2, I3) [0 <= I0 - 2 * I2 /\ I0 - 2 * I2 <= 1 /\ 1 <= I0 - 1 /\ I2 <= I0 - 1]
0.00/0.52	  f2#(I4, I5) -> f3#(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.52	R = 
0.00/0.52	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.52	  f3(I0, I1) -> f2(I2, I3) [0 <= I0 - 2 * I2 /\ I0 - 2 * I2 <= 1 /\ 1 <= I0 - 1 /\ I2 <= I0 - 1]
0.00/0.52	  f2(I4, I5) -> f3(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.52	
0.00/0.52	The dependency graph for this problem is:
0.00/0.52	  0 -> 3
0.00/0.52	  1 -> 2
0.00/0.52	  2 -> 1
0.00/0.52	  3 -> 2
0.00/0.52	Where:
0.00/0.52	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.52	  1) f3#(I0, I1) -> f2#(I2, I3) [0 <= I0 - 2 * I2 /\ I0 - 2 * I2 <= 1 /\ 1 <= I0 - 1 /\ I2 <= I0 - 1]
0.00/0.52	  2) f2#(I4, I5) -> f3#(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	  3) f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.52	
0.00/0.52	We have the following SCCs.
0.00/0.52	  { 1, 2 }
0.00/0.52	
0.00/0.52	DP problem for innermost termination.
0.00/0.52	P =
0.00/0.52	  f3#(I0, I1) -> f2#(I2, I3) [0 <= I0 - 2 * I2 /\ I0 - 2 * I2 <= 1 /\ 1 <= I0 - 1 /\ I2 <= I0 - 1]
0.00/0.52	  f2#(I4, I5) -> f3#(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	R = 
0.00/0.52	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.52	  f3(I0, I1) -> f2(I2, I3) [0 <= I0 - 2 * I2 /\ I0 - 2 * I2 <= 1 /\ 1 <= I0 - 1 /\ I2 <= I0 - 1]
0.00/0.52	  f2(I4, I5) -> f3(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.52	
0.00/0.52	We use the basic value criterion with the projection function NU:
0.00/0.52	  NU[f2#(z1,z2)] = z1
0.00/0.52	  NU[f3#(z1,z2)] = z1
0.00/0.52	
0.00/0.52	This gives the following inequalities:
0.00/0.52	  0 <= I0 - 2 * I2 /\ I0 - 2 * I2 <= 1 /\ 1 <= I0 - 1 /\ I2 <= I0 - 1  ==>  I0 >! I2
0.00/0.52	  1 <= I4 - 1 /\ y1 <= I4 - 1  ==>  I4 (>! \union =) I4
0.00/0.52	
0.00/0.52	We remove all the strictly oriented dependency pairs.
0.00/0.52	
0.00/0.52	DP problem for innermost termination.
0.00/0.52	P =
0.00/0.52	  f2#(I4, I5) -> f3#(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	R = 
0.00/0.52	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.52	  f3(I0, I1) -> f2(I2, I3) [0 <= I0 - 2 * I2 /\ I0 - 2 * I2 <= 1 /\ 1 <= I0 - 1 /\ I2 <= I0 - 1]
0.00/0.52	  f2(I4, I5) -> f3(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.52	
0.00/0.52	The dependency graph for this problem is:
0.00/0.52	  2 -> 
0.00/0.52	Where:
0.00/0.52	  2) f2#(I4, I5) -> f3#(I4, I6) [1 <= I4 - 1 /\ y1 <= I4 - 1]
0.00/0.52	
0.00/0.52	We have the following SCCs.
0.00/0.52	  
0.00/3.50	EOF