0.51/0.63	YES
0.51/0.63	
0.51/0.63	DP problem for innermost termination.
0.51/0.63	P =
0.51/0.63	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.51/0.63	  f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 2) [0 <= I0 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1 /\ -1 <= y1 - 1 /\ -1 <= I3 - 1 /\ y1 <= 41]
0.51/0.63	  f2#(I4, I5, I6) -> f2#(I4 - 1, I5, I6 + 1) [-1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1]
0.51/0.63	  f1#(I8, I9, I10) -> f2#(I11, I12, 2) [0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I11 - 1]
0.51/0.63	R = 
0.51/0.63	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.51/0.63	  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 2) [0 <= I0 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1 /\ -1 <= y1 - 1 /\ -1 <= I3 - 1 /\ y1 <= 41]
0.51/0.63	  f2(I4, I5, I6) -> f2(I4 - 1, I5, I6 + 1) [-1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1]
0.51/0.63	  f1(I8, I9, I10) -> f2(I11, I12, 2) [0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I11 - 1]
0.51/0.63	
0.51/0.63	The dependency graph for this problem is:
0.51/0.63	  0 -> 3
0.51/0.63	  1 -> 1, 2
0.51/0.63	  2 -> 1, 2
0.51/0.63	  3 -> 1, 2
0.51/0.63	Where:
0.51/0.63	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.51/0.63	  1) f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 2) [0 <= I0 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1 /\ -1 <= y1 - 1 /\ -1 <= I3 - 1 /\ y1 <= 41]
0.51/0.63	  2) f2#(I4, I5, I6) -> f2#(I4 - 1, I5, I6 + 1) [-1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1]
0.51/0.63	  3) f1#(I8, I9, I10) -> f2#(I11, I12, 2) [0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I11 - 1]
0.51/0.63	
0.51/0.63	We have the following SCCs.
0.51/0.63	  { 1, 2 }
0.51/0.63	
0.51/0.63	DP problem for innermost termination.
0.51/0.63	P =
0.51/0.63	  f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 2) [0 <= I0 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1 /\ -1 <= y1 - 1 /\ -1 <= I3 - 1 /\ y1 <= 41]
0.51/0.63	  f2#(I4, I5, I6) -> f2#(I4 - 1, I5, I6 + 1) [-1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1]
0.51/0.63	R = 
0.51/0.63	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.51/0.63	  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 2) [0 <= I0 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1 /\ -1 <= y1 - 1 /\ -1 <= I3 - 1 /\ y1 <= 41]
0.51/0.63	  f2(I4, I5, I6) -> f2(I4 - 1, I5, I6 + 1) [-1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1]
0.51/0.63	  f1(I8, I9, I10) -> f2(I11, I12, 2) [0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I11 - 1]
0.51/0.63	
0.51/0.63	We use the basic value criterion with the projection function NU:
0.51/0.63	  NU[f2#(z1,z2,z3)] = z2
0.51/0.63	
0.51/0.63	This gives the following inequalities:
0.51/0.63	  0 <= I0 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1 /\ -1 <= y1 - 1 /\ -1 <= I3 - 1 /\ y1 <= 41  ==>  I1 >! I1 - 1
0.51/0.63	  -1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I5 (>! \union =) I5
0.51/0.63	
0.51/0.63	We remove all the strictly oriented dependency pairs.
0.51/0.63	
0.51/0.63	DP problem for innermost termination.
0.51/0.63	P =
0.51/0.63	  f2#(I4, I5, I6) -> f2#(I4 - 1, I5, I6 + 1) [-1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1]
0.51/0.63	R = 
0.51/0.63	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.51/0.63	  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 2) [0 <= I0 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1 /\ -1 <= y1 - 1 /\ -1 <= I3 - 1 /\ y1 <= 41]
0.51/0.63	  f2(I4, I5, I6) -> f2(I4 - 1, I5, I6 + 1) [-1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1]
0.51/0.63	  f1(I8, I9, I10) -> f2(I11, I12, 2) [0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I11 - 1]
0.51/0.63	
0.51/0.63	We use the basic value criterion with the projection function NU:
0.51/0.63	  NU[f2#(z1,z2,z3)] = z1
0.51/0.63	
0.51/0.63	This gives the following inequalities:
0.51/0.63	  -1 <= I6 - 1 /\ 41 <= I7 - 1 /\ 0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I4 >! I4 - 1
0.51/0.63	
0.51/0.63	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.51/3.62	EOF