0.00/0.58	YES
0.00/0.58	
0.00/0.58	DP problem for innermost termination.
0.00/0.58	P =
0.00/0.58	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.58	  f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
0.00/0.58	  f2#(I2, I3) -> f2#(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
0.00/0.58	  f1#(I4, I5) -> f2#(0, 10) 
0.00/0.58	R = 
0.00/0.58	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.58	  f2(I0, I1) -> f2(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
0.00/0.58	  f2(I2, I3) -> f2(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
0.00/0.58	  f1(I4, I5) -> f2(0, 10) 
0.00/0.58	
0.00/0.58	The dependency graph for this problem is:
0.00/0.58	  0 -> 3
0.00/0.58	  1 -> 1, 2
0.00/0.58	  2 -> 1, 2
0.00/0.58	  3 -> 1, 2
0.00/0.58	Where:
0.00/0.58	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.58	  1) f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
0.00/0.58	  2) f2#(I2, I3) -> f2#(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
0.00/0.58	  3) f1#(I4, I5) -> f2#(0, 10) 
0.00/0.58	
0.00/0.58	We have the following SCCs.
0.00/0.58	  { 1, 2 }
0.00/0.58	
0.00/0.58	DP problem for innermost termination.
0.00/0.58	P =
0.00/0.58	  f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
0.00/0.58	  f2#(I2, I3) -> f2#(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
0.00/0.58	R = 
0.00/0.58	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.58	  f2(I0, I1) -> f2(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
0.00/0.58	  f2(I2, I3) -> f2(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
0.00/0.58	  f1(I4, I5) -> f2(0, 10) 
0.00/0.58	
0.00/0.58	We use the basic value criterion with the projection function NU:
0.00/0.58	  NU[f2#(z1,z2)] = z2
0.00/0.58	
0.00/0.58	This gives the following inequalities:
0.00/0.58	  0 <= I1 - 1 /\ I0 <= I1 - 1  ==>  I1 (>! \union =) I1
0.00/0.58	  0 <= I3 - 1 /\ I2 <= I3 - 1  ==>  I3 >! I2
0.00/0.58	
0.00/0.58	We remove all the strictly oriented dependency pairs.
0.00/0.58	
0.00/0.58	DP problem for innermost termination.
0.00/0.58	P =
0.00/0.58	  f2#(I0, I1) -> f2#(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
0.00/0.58	R = 
0.00/0.58	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.58	  f2(I0, I1) -> f2(I0 + 1, I1) [0 <= I1 - 1 /\ I0 <= I1 - 1]
0.00/0.58	  f2(I2, I3) -> f2(0, I2) [0 <= I3 - 1 /\ I2 <= I3 - 1]
0.00/0.58	  f1(I4, I5) -> f2(0, 10) 
0.00/0.58	
0.00/0.58	We use the reverse value criterion with the projection function NU:
0.00/0.58	  NU[f2#(z1,z2)] = z2 - 1 + -1 * z1
0.00/0.58	
0.00/0.58	This gives the following inequalities:
0.00/0.58	  0 <= I1 - 1 /\ I0 <= I1 - 1  ==>  I1 - 1 + -1 * I0 > I1 - 1 + -1 * (I0 + 1)  with  I1 - 1 + -1 * I0 >= 0
0.00/0.58	
0.00/0.58	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.56	EOF