0.55/0.59	YES
0.55/0.59	
0.55/0.59	DP problem for innermost termination.
0.55/0.59	P =
0.55/0.59	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.55/0.59	  f2#(I0, I1) -> f2#(I2, I1 - 1) [I1 - 1 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.55/0.59	  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.55/0.59	  f2#(I5, I6) -> f2#(1, I6 - 1) [0 = I5 /\ I6 - 1 <= I6 - 1 /\ 0 <= I6 - 1]
0.55/0.59	  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I10 - 1 /\ 1 <= I8 - 1 /\ -1 <= I9 - 1]
0.55/0.59	  f1#(I11, I12) -> f2#(0, I13) [1 = I12 /\ -1 <= I13 - 1 /\ 0 <= I11 - 1]
0.55/0.59	  f1#(I14, I15) -> f2#(0, 0) [0 = I15 /\ 0 <= I14 - 1]
0.55/0.59	R = 
0.55/0.59	  init(x1, x2) -> f1(rnd1, rnd2) 
0.55/0.59	  f2(I0, I1) -> f2(I2, I1 - 1) [I1 - 1 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.55/0.59	  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.55/0.59	  f2(I5, I6) -> f2(1, I6 - 1) [0 = I5 /\ I6 - 1 <= I6 - 1 /\ 0 <= I6 - 1]
0.55/0.59	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I10 - 1 /\ 1 <= I8 - 1 /\ -1 <= I9 - 1]
0.55/0.59	  f1(I11, I12) -> f2(0, I13) [1 = I12 /\ -1 <= I13 - 1 /\ 0 <= I11 - 1]
0.55/0.59	  f1(I14, I15) -> f2(0, 0) [0 = I15 /\ 0 <= I14 - 1]
0.55/0.59	
0.55/0.59	The dependency graph for this problem is:
0.55/0.59	  0 -> 4, 5, 6
0.55/0.59	  1 -> 1, 2
0.55/0.59	  2 -> 1, 2, 3
0.55/0.59	  3 -> 1, 2
0.55/0.59	  4 -> 1, 2, 3
0.55/0.59	  5 -> 3
0.55/0.59	  6 -> 
0.55/0.59	Where:
0.55/0.59	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.55/0.59	  1) f2#(I0, I1) -> f2#(I2, I1 - 1) [I1 - 1 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.55/0.59	  2) f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.55/0.59	  3) f2#(I5, I6) -> f2#(1, I6 - 1) [0 = I5 /\ I6 - 1 <= I6 - 1 /\ 0 <= I6 - 1]
0.55/0.59	  4) f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I10 - 1 /\ 1 <= I8 - 1 /\ -1 <= I9 - 1]
0.55/0.59	  5) f1#(I11, I12) -> f2#(0, I13) [1 = I12 /\ -1 <= I13 - 1 /\ 0 <= I11 - 1]
0.55/0.59	  6) f1#(I14, I15) -> f2#(0, 0) [0 = I15 /\ 0 <= I14 - 1]
0.55/0.59	
0.55/0.59	We have the following SCCs.
0.55/0.59	  { 1, 2, 3 }
0.55/0.59	
0.55/0.59	DP problem for innermost termination.
0.55/0.59	P =
0.55/0.59	  f2#(I0, I1) -> f2#(I2, I1 - 1) [I1 - 1 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.55/0.59	  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.55/0.59	  f2#(I5, I6) -> f2#(1, I6 - 1) [0 = I5 /\ I6 - 1 <= I6 - 1 /\ 0 <= I6 - 1]
0.55/0.59	R = 
0.55/0.59	  init(x1, x2) -> f1(rnd1, rnd2) 
0.55/0.59	  f2(I0, I1) -> f2(I2, I1 - 1) [I1 - 1 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.55/0.59	  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.55/0.59	  f2(I5, I6) -> f2(1, I6 - 1) [0 = I5 /\ I6 - 1 <= I6 - 1 /\ 0 <= I6 - 1]
0.55/0.59	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I10 - 1 /\ 1 <= I8 - 1 /\ -1 <= I9 - 1]
0.55/0.59	  f1(I11, I12) -> f2(0, I13) [1 = I12 /\ -1 <= I13 - 1 /\ 0 <= I11 - 1]
0.55/0.59	  f1(I14, I15) -> f2(0, 0) [0 = I15 /\ 0 <= I14 - 1]
0.55/0.59	
0.55/0.59	We use the basic value criterion with the projection function NU:
0.55/0.59	  NU[f2#(z1,z2)] = z2
0.55/0.59	
0.55/0.59	This gives the following inequalities:
0.55/0.59	  I1 - 1 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1  ==>  I1 >! I1 - 1
0.55/0.59	  0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1  ==>  I4 (>! \union =) I4
0.55/0.59	  0 = I5 /\ I6 - 1 <= I6 - 1 /\ 0 <= I6 - 1  ==>  I6 >! I6 - 1
0.55/0.59	
0.55/0.59	We remove all the strictly oriented dependency pairs.
0.55/0.59	
0.55/0.59	DP problem for innermost termination.
0.55/0.59	P =
0.55/0.59	  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.55/0.59	R = 
0.55/0.59	  init(x1, x2) -> f1(rnd1, rnd2) 
0.55/0.59	  f2(I0, I1) -> f2(I2, I1 - 1) [I1 - 1 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1]
0.55/0.59	  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.55/0.59	  f2(I5, I6) -> f2(1, I6 - 1) [0 = I5 /\ I6 - 1 <= I6 - 1 /\ 0 <= I6 - 1]
0.55/0.59	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I10 - 1 /\ 1 <= I8 - 1 /\ -1 <= I9 - 1]
0.55/0.59	  f1(I11, I12) -> f2(0, I13) [1 = I12 /\ -1 <= I13 - 1 /\ 0 <= I11 - 1]
0.55/0.59	  f1(I14, I15) -> f2(0, 0) [0 = I15 /\ 0 <= I14 - 1]
0.55/0.59	
0.55/0.59	We use the basic value criterion with the projection function NU:
0.55/0.59	  NU[f2#(z1,z2)] = z1
0.55/0.59	
0.55/0.59	This gives the following inequalities:
0.55/0.59	  0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1  ==>  I3 >! I3 - 1
0.55/0.59	
0.55/0.59	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.55/3.58	EOF