0.00/0.41	YES
0.00/0.41	
0.00/0.41	DP problem for innermost termination.
0.00/0.41	P =
0.00/0.41	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.41	  f2#(I0, I1) -> f2#(I2, I1 + 1) [I1 + 2 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 - 1 <= I0 /\ -1 <= I1 - 1 /\ I1 <= 9]
0.00/0.41	  f1#(I3, I4) -> f2#(I5, 0) [1 <= I5 - 1]
0.00/0.41	R = 
0.00/0.41	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.41	  f2(I0, I1) -> f2(I2, I1 + 1) [I1 + 2 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 - 1 <= I0 /\ -1 <= I1 - 1 /\ I1 <= 9]
0.00/0.41	  f1(I3, I4) -> f2(I5, 0) [1 <= I5 - 1]
0.00/0.41	
0.00/0.41	The dependency graph for this problem is:
0.00/0.41	  0 -> 2
0.00/0.41	  1 -> 1
0.00/0.41	  2 -> 1
0.00/0.41	Where:
0.00/0.41	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.41	  1) f2#(I0, I1) -> f2#(I2, I1 + 1) [I1 + 2 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 - 1 <= I0 /\ -1 <= I1 - 1 /\ I1 <= 9]
0.00/0.41	  2) f1#(I3, I4) -> f2#(I5, 0) [1 <= I5 - 1]
0.00/0.41	
0.00/0.41	We have the following SCCs.
0.00/0.41	  { 1 }
0.00/0.41	
0.00/0.41	DP problem for innermost termination.
0.00/0.41	P =
0.00/0.41	  f2#(I0, I1) -> f2#(I2, I1 + 1) [I1 + 2 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 - 1 <= I0 /\ -1 <= I1 - 1 /\ I1 <= 9]
0.00/0.41	R = 
0.00/0.41	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.41	  f2(I0, I1) -> f2(I2, I1 + 1) [I1 + 2 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 - 1 <= I0 /\ -1 <= I1 - 1 /\ I1 <= 9]
0.00/0.41	  f1(I3, I4) -> f2(I5, 0) [1 <= I5 - 1]
0.00/0.41	
0.00/0.41	We use the reverse value criterion with the projection function NU:
0.00/0.41	  NU[f2#(z1,z2)] = 9 + -1 * z2
0.00/0.41	
0.00/0.41	This gives the following inequalities:
0.00/0.41	  I1 + 2 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 - 1 <= I0 /\ -1 <= I1 - 1 /\ I1 <= 9  ==>  9 + -1 * I1 > 9 + -1 * (I1 + 1)  with  9 + -1 * I1 >= 0
0.00/0.41	
0.00/0.41	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.38	EOF