2.26/2.60	MAYBE
2.26/2.60	
2.26/2.60	DP problem for innermost termination.
2.26/2.60	P =
2.26/2.60	  init#(x1, x2) -> f1#(rnd1, rnd2) 
2.26/2.60	  f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1]
2.26/2.60	  f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1]
2.26/2.60	  f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10]
2.26/2.60	  f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1]
2.26/2.60	  f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
2.26/2.60	R = 
2.26/2.60	  init(x1, x2) -> f1(rnd1, rnd2) 
2.26/2.60	  f3(I0, I1) -> f2(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1]
2.26/2.60	  f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1]
2.26/2.60	  f3(I7, I8) -> f2(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10]
2.26/2.60	  f2(I11, I12) -> f3(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1]
2.26/2.60	  f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
2.26/2.60	
2.26/2.60	The dependency graph for this problem is:
2.26/2.60	  0 -> 5
2.26/2.60	  1 -> 2, 4
2.26/2.60	  2 -> 1, 3
2.26/2.60	  3 -> 2, 4
2.26/2.60	  4 -> 3
2.26/2.60	  5 -> 2, 4
2.26/2.60	Where:
2.26/2.60	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
2.26/2.60	  1) f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1]
2.26/2.60	  2) f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1]
2.26/2.60	  3) f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10]
2.26/2.60	  4) f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1]
2.26/2.60	  5) f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
2.26/2.60	
2.26/2.60	We have the following SCCs.
2.26/2.60	  { 1, 2, 3, 4 }
2.26/2.60	
2.26/2.60	DP problem for innermost termination.
2.26/2.60	P =
2.26/2.60	  f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1]
2.26/2.60	  f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1]
2.26/2.60	  f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10]
2.26/2.60	  f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1]
2.26/2.60	R = 
2.26/2.60	  init(x1, x2) -> f1(rnd1, rnd2) 
2.26/2.60	  f3(I0, I1) -> f2(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1]
2.26/2.60	  f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1]
2.26/2.60	  f3(I7, I8) -> f2(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10]
2.26/2.60	  f2(I11, I12) -> f3(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1]
2.26/2.60	  f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
2.26/2.60	
2.26/2.60	We use the basic value criterion with the projection function NU:
2.26/2.60	  NU[f2#(z1,z2)] = z1
2.26/2.60	  NU[f3#(z1,z2)] = z1
2.26/2.60	
2.26/2.60	This gives the following inequalities:
2.26/2.60	  0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1  ==>  I0 >! I0 - 1
2.26/2.60	  0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1  ==>  I3 (>! \union =) I3
2.26/2.60	  2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10  ==>  I7 (>! \union =) I7
2.26/2.60	  I11 - 5 * I14 = 0 /\ 2 <= I11 - 1  ==>  I11 (>! \union =) I11
2.26/2.60	
2.26/2.60	We remove all the strictly oriented dependency pairs.
2.26/2.60	
2.26/2.60	DP problem for innermost termination.
2.26/2.60	P =
2.26/2.60	  f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1]
2.26/2.60	  f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10]
2.26/2.60	  f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1]
2.26/2.60	R = 
2.26/2.60	  init(x1, x2) -> f1(rnd1, rnd2) 
2.26/2.60	  f3(I0, I1) -> f2(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1]
2.26/2.60	  f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1]
2.26/2.60	  f3(I7, I8) -> f2(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10]
2.26/2.60	  f2(I11, I12) -> f3(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1]
2.26/2.60	  f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
2.26/2.60	
2.26/5.58	EOF