0.53/0.62	MAYBE
0.53/0.62	
0.53/0.62	DP problem for innermost termination.
0.53/0.62	P =
0.53/0.62	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.53/0.62	  f2#(I0, I1) -> f2#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1]
0.53/0.62	  f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1]
0.53/0.62	  f2#(I6, I7) -> f2#(-5, I8) [-5 = I6]
0.53/0.62	  f1#(I9, I10) -> f2#(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0]
0.53/0.62	R = 
0.53/0.62	  init(x1, x2) -> f1(rnd1, rnd2) 
0.53/0.62	  f2(I0, I1) -> f2(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1]
0.53/0.62	  f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1]
0.53/0.62	  f2(I6, I7) -> f2(-5, I8) [-5 = I6]
0.53/0.62	  f1(I9, I10) -> f2(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0]
0.53/0.62	
0.53/0.62	The dependency graph for this problem is:
0.53/0.62	  0 -> 4
0.53/0.62	  1 -> 1
0.53/0.62	  2 -> 2, 3
0.53/0.62	  3 -> 3
0.53/0.62	  4 -> 1, 2, 3
0.53/0.62	Where:
0.53/0.62	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.53/0.62	  1) f2#(I0, I1) -> f2#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1]
0.53/0.62	  2) f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1]
0.53/0.62	  3) f2#(I6, I7) -> f2#(-5, I8) [-5 = I6]
0.53/0.62	  4) f1#(I9, I10) -> f2#(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0]
0.53/0.62	
0.53/0.62	We have the following SCCs.
0.53/0.62	  { 2 }
0.53/0.62	  { 3 }
0.53/0.62	  { 1 }
0.53/0.62	
0.53/0.62	DP problem for innermost termination.
0.53/0.62	P =
0.53/0.62	  f2#(I0, I1) -> f2#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1]
0.53/0.62	R = 
0.53/0.62	  init(x1, x2) -> f1(rnd1, rnd2) 
0.53/0.62	  f2(I0, I1) -> f2(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1]
0.53/0.62	  f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1]
0.53/0.62	  f2(I6, I7) -> f2(-5, I8) [-5 = I6]
0.53/0.62	  f1(I9, I10) -> f2(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0]
0.53/0.62	
0.53/0.62	We use the reverse value criterion with the projection function NU:
0.53/0.62	  NU[f2#(z1,z2)] = 0 + -1 * z1
0.53/0.62	
0.53/0.62	This gives the following inequalities:
0.53/0.62	  -5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1  ==>  0 + -1 * I0 > 0 + -1 * (I0 + 1)  with  0 + -1 * I0 >= 0
0.53/0.62	
0.53/0.62	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.53/0.62	
0.53/0.62	DP problem for innermost termination.
0.53/0.62	P =
0.53/0.62	  f2#(I6, I7) -> f2#(-5, I8) [-5 = I6]
0.53/0.62	R = 
0.53/0.62	  init(x1, x2) -> f1(rnd1, rnd2) 
0.53/0.62	  f2(I0, I1) -> f2(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1]
0.53/0.62	  f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1]
0.53/0.62	  f2(I6, I7) -> f2(-5, I8) [-5 = I6]
0.53/0.62	  f1(I9, I10) -> f2(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0]
0.53/0.62	
0.62/3.60	EOF