0.00/0.38	YES
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  f3#(I0, I1) -> f3#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
0.00/0.38	  f2#(I4, I5) -> f2#(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4]
0.00/0.38	  f2#(I7, I8) -> f3#(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7]
0.00/0.38	  f1#(I11, I12) -> f2#(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f3(I0, I1) -> f3(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
0.00/0.38	  f2(I4, I5) -> f2(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4]
0.00/0.38	  f2(I7, I8) -> f3(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7]
0.00/0.38	  f1(I11, I12) -> f2(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14]
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  0 -> 4
0.00/0.38	  1 -> 1
0.00/0.38	  2 -> 2, 3
0.00/0.38	  3 -> 1
0.00/0.38	  4 -> 2, 3
0.00/0.38	Where:
0.00/0.38	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  1) f3#(I0, I1) -> f3#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
0.00/0.38	  2) f2#(I4, I5) -> f2#(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4]
0.00/0.38	  3) f2#(I7, I8) -> f3#(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7]
0.00/0.38	  4) f1#(I11, I12) -> f2#(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14]
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  { 2 }
0.00/0.38	  { 1 }
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f3#(I0, I1) -> f3#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f3(I0, I1) -> f3(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
0.00/0.38	  f2(I4, I5) -> f2(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4]
0.00/0.38	  f2(I7, I8) -> f3(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7]
0.00/0.38	  f1(I11, I12) -> f2(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14]
0.00/0.38	
0.00/0.38	We use the basic value criterion with the projection function NU:
0.00/0.38	  NU[f3#(z1,z2)] = z1
0.00/0.38	
0.00/0.38	This gives the following inequalities:
0.00/0.38	  -1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0  ==>  I0 >! I2
0.00/0.38	
0.00/0.38	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I4, I5) -> f2#(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f3(I0, I1) -> f3(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0]
0.00/0.38	  f2(I4, I5) -> f2(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4]
0.00/0.38	  f2(I7, I8) -> f3(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7]
0.00/0.38	  f1(I11, I12) -> f2(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14]
0.00/0.38	
0.00/0.38	We use the basic value criterion with the projection function NU:
0.00/0.38	  NU[f2#(z1,z2)] = z2
0.00/0.38	
0.00/0.38	This gives the following inequalities:
0.00/0.38	  0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4  ==>  I5 >! I5 - 1
0.00/0.38	
0.00/0.38	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.37	EOF