0.00/0.67	MAYBE
0.00/0.67	
0.00/0.67	DP problem for innermost termination.
0.00/0.67	P =
0.00/0.67	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.67	  f3#(I0, I1) -> f3#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= -1]
0.00/0.67	  f3#(I3, I4) -> f3#(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= -1]
0.00/0.67	  f3#(I6, I7) -> f3#(-5, I8) [-5 = I6]
0.00/0.67	  f2#(I9, I10) -> f3#(I11, I12) [-1 <= y1 - 1 /\ 1 <= I10 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I9 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I11]
0.00/0.67	  f1#(I13, I14) -> f2#(I13, I14) [-1 <= I15 - 1 /\ 1 <= I14 - 1 /\ I16 - 2 * I17 = 0 /\ -1 <= I16 - 1 /\ 0 <= I13 - 1]
0.00/0.67	R = 
0.00/0.67	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.67	  f3(I0, I1) -> f3(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= -1]
0.00/0.67	  f3(I3, I4) -> f3(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= -1]
0.00/0.67	  f3(I6, I7) -> f3(-5, I8) [-5 = I6]
0.00/0.67	  f2(I9, I10) -> f3(I11, I12) [-1 <= y1 - 1 /\ 1 <= I10 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I9 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I11]
0.00/0.67	  f1(I13, I14) -> f2(I13, I14) [-1 <= I15 - 1 /\ 1 <= I14 - 1 /\ I16 - 2 * I17 = 0 /\ -1 <= I16 - 1 /\ 0 <= I13 - 1]
0.00/0.67	
0.00/0.67	The dependency graph for this problem is:
0.00/0.67	  0 -> 5
0.00/0.67	  1 -> 1
0.00/0.67	  2 -> 2, 3
0.00/0.67	  3 -> 3
0.00/0.67	  4 -> 1, 2, 3
0.00/0.67	  5 -> 4
0.00/0.67	Where:
0.00/0.67	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.67	  1) f3#(I0, I1) -> f3#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= -1]
0.00/0.67	  2) f3#(I3, I4) -> f3#(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= -1]
0.00/0.67	  3) f3#(I6, I7) -> f3#(-5, I8) [-5 = I6]
0.00/0.67	  4) f2#(I9, I10) -> f3#(I11, I12) [-1 <= y1 - 1 /\ 1 <= I10 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I9 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I11]
0.00/0.67	  5) f1#(I13, I14) -> f2#(I13, I14) [-1 <= I15 - 1 /\ 1 <= I14 - 1 /\ I16 - 2 * I17 = 0 /\ -1 <= I16 - 1 /\ 0 <= I13 - 1]
0.00/0.67	
0.00/0.67	We have the following SCCs.
0.00/0.67	  { 2 }
0.00/0.67	  { 3 }
0.00/0.67	  { 1 }
0.00/0.67	
0.00/0.67	DP problem for innermost termination.
0.00/0.67	P =
0.00/0.67	  f3#(I0, I1) -> f3#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= -1]
0.00/0.67	R = 
0.00/0.67	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.67	  f3(I0, I1) -> f3(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= -1]
0.00/0.67	  f3(I3, I4) -> f3(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= -1]
0.00/0.67	  f3(I6, I7) -> f3(-5, I8) [-5 = I6]
0.00/0.67	  f2(I9, I10) -> f3(I11, I12) [-1 <= y1 - 1 /\ 1 <= I10 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I9 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I11]
0.00/0.67	  f1(I13, I14) -> f2(I13, I14) [-1 <= I15 - 1 /\ 1 <= I14 - 1 /\ I16 - 2 * I17 = 0 /\ -1 <= I16 - 1 /\ 0 <= I13 - 1]
0.00/0.67	
0.00/0.67	We use the reverse value criterion with the projection function NU:
0.00/0.67	  NU[f3#(z1,z2)] = -1 + -1 * z1
0.00/0.67	
0.00/0.67	This gives the following inequalities:
0.00/0.67	  -5 <= I0 - 1 /\ I0 <= -1  ==>  -1 + -1 * I0 > -1 + -1 * (I0 + 1)  with  -1 + -1 * I0 >= 0
0.00/0.67	
0.00/0.67	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.67	
0.00/0.67	DP problem for innermost termination.
0.00/0.67	P =
0.00/0.67	  f3#(I6, I7) -> f3#(-5, I8) [-5 = I6]
0.00/0.67	R = 
0.00/0.67	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.67	  f3(I0, I1) -> f3(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= -1]
0.00/0.67	  f3(I3, I4) -> f3(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= -1]
0.00/0.67	  f3(I6, I7) -> f3(-5, I8) [-5 = I6]
0.00/0.67	  f2(I9, I10) -> f3(I11, I12) [-1 <= y1 - 1 /\ 1 <= I10 - 1 /\ y2 - 2 * y3 = 0 /\ -1 <= y2 - 1 /\ 0 <= I9 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I11]
0.00/0.67	  f1(I13, I14) -> f2(I13, I14) [-1 <= I15 - 1 /\ 1 <= I14 - 1 /\ I16 - 2 * I17 = 0 /\ -1 <= I16 - 1 /\ 0 <= I13 - 1]
0.00/0.67	
0.00/3.65	EOF