0.00/0.42	YES
0.00/0.42	
0.00/0.42	DP problem for innermost termination.
0.00/0.42	P =
0.00/0.42	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.42	  f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.42	  f2#(I2, I3) -> f2#(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.00/0.42	  f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.42	R = 
0.00/0.42	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.42	  f2(I0, I1) -> f2(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.42	  f2(I2, I3) -> f2(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.00/0.42	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.42	
0.00/0.42	The dependency graph for this problem is:
0.00/0.42	  0 -> 3
0.00/0.42	  1 -> 1, 2
0.00/0.42	  2 -> 1, 2
0.00/0.42	  3 -> 1, 2
0.00/0.42	Where:
0.00/0.42	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.42	  1) f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.42	  2) f2#(I2, I3) -> f2#(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.00/0.42	  3) f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.42	
0.00/0.42	We have the following SCCs.
0.00/0.42	  { 1, 2 }
0.00/0.42	
0.00/0.42	DP problem for innermost termination.
0.00/0.42	P =
0.00/0.42	  f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.42	  f2#(I2, I3) -> f2#(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.00/0.42	R = 
0.00/0.42	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.42	  f2(I0, I1) -> f2(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.42	  f2(I2, I3) -> f2(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.00/0.42	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.42	
0.00/0.42	We use the basic value criterion with the projection function NU:
0.00/0.42	  NU[f2#(z1,z2)] = z2
0.00/0.42	
0.00/0.42	This gives the following inequalities:
0.00/0.42	  I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1  ==>  I1 (>! \union =) I1
0.00/0.42	  I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1  ==>  I3 >! I3 - I2
0.00/0.42	
0.00/0.42	We remove all the strictly oriented dependency pairs.
0.00/0.42	
0.00/0.42	DP problem for innermost termination.
0.00/0.42	P =
0.00/0.42	  f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.42	R = 
0.00/0.42	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.42	  f2(I0, I1) -> f2(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.42	  f2(I2, I3) -> f2(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1]
0.00/0.42	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.42	
0.00/0.42	We use the basic value criterion with the projection function NU:
0.00/0.42	  NU[f2#(z1,z2)] = z1
0.00/0.42	
0.00/0.42	This gives the following inequalities:
0.00/0.42	  I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1  ==>  I0 >! I0 - I1
0.00/0.42	
0.00/0.42	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.40	EOF