0.00/0.29	YES
0.00/0.29	
0.00/0.29	DP problem for innermost termination.
0.00/0.29	P =
0.00/0.29	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.29	  f2#(I0, I1) -> f2#(I0 - 1, I1) [1 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 1]
0.00/0.29	  f2#(I2, I3) -> f2#(I2 - 1, I3) [1 <= I2 - 1 /\ I3 <= 1 /\ I2 - 1 <= I2 - 1]
0.00/0.29	  f1#(I4, I5) -> f2#(I6, 1) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.29	  f1#(I7, I8) -> f2#(0, 0) [0 = I8 /\ 0 <= I7 - 1]
0.00/0.29	R = 
0.00/0.29	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.29	  f2(I0, I1) -> f2(I0 - 1, I1) [1 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 1]
0.00/0.29	  f2(I2, I3) -> f2(I2 - 1, I3) [1 <= I2 - 1 /\ I3 <= 1 /\ I2 - 1 <= I2 - 1]
0.00/0.29	  f1(I4, I5) -> f2(I6, 1) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.29	  f1(I7, I8) -> f2(0, 0) [0 = I8 /\ 0 <= I7 - 1]
0.00/0.29	
0.00/0.29	The dependency graph for this problem is:
0.00/0.29	  0 -> 3, 4
0.00/0.29	  1 -> 1, 2
0.00/0.29	  2 -> 1, 2
0.00/0.29	  3 -> 1, 2
0.00/0.29	  4 -> 
0.00/0.29	Where:
0.00/0.29	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.29	  1) f2#(I0, I1) -> f2#(I0 - 1, I1) [1 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 1]
0.00/0.29	  2) f2#(I2, I3) -> f2#(I2 - 1, I3) [1 <= I2 - 1 /\ I3 <= 1 /\ I2 - 1 <= I2 - 1]
0.00/0.29	  3) f1#(I4, I5) -> f2#(I6, 1) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.29	  4) f1#(I7, I8) -> f2#(0, 0) [0 = I8 /\ 0 <= I7 - 1]
0.00/0.29	
0.00/0.29	We have the following SCCs.
0.00/0.29	  { 1, 2 }
0.00/0.29	
0.00/0.29	DP problem for innermost termination.
0.00/0.29	P =
0.00/0.29	  f2#(I0, I1) -> f2#(I0 - 1, I1) [1 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 1]
0.00/0.29	  f2#(I2, I3) -> f2#(I2 - 1, I3) [1 <= I2 - 1 /\ I3 <= 1 /\ I2 - 1 <= I2 - 1]
0.00/0.29	R = 
0.00/0.29	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.29	  f2(I0, I1) -> f2(I0 - 1, I1) [1 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 1]
0.00/0.29	  f2(I2, I3) -> f2(I2 - 1, I3) [1 <= I2 - 1 /\ I3 <= 1 /\ I2 - 1 <= I2 - 1]
0.00/0.29	  f1(I4, I5) -> f2(I6, 1) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.29	  f1(I7, I8) -> f2(0, 0) [0 = I8 /\ 0 <= I7 - 1]
0.00/0.29	
0.00/0.29	We use the basic value criterion with the projection function NU:
0.00/0.29	  NU[f2#(z1,z2)] = z1
0.00/0.29	
0.00/0.29	This gives the following inequalities:
0.00/0.29	  1 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 1  ==>  I0 >! I0 - 1
0.00/0.29	  1 <= I2 - 1 /\ I3 <= 1 /\ I2 - 1 <= I2 - 1  ==>  I2 >! I2 - 1
0.00/0.29	
0.00/0.29	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.27	EOF