0.99/1.07	MAYBE
0.99/1.07	
0.99/1.07	DP problem for innermost termination.
0.99/1.07	P =
0.99/1.07	  init#(x1, x2) -> f3#(rnd1, rnd2) 
0.99/1.07	  f5#(I0, I1) -> f5#(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.99/1.07	  f3#(I2, I3) -> f5#(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.99/1.07	  f4#(I5, I6) -> f4#(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1]
0.99/1.07	  f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1]
0.99/1.07	  f4#(I14, I15) -> f4#(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1]
0.99/1.07	  f2#(I19, I20) -> f4#(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1]
0.99/1.07	  f3#(I24, I25) -> f2#(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1]
0.99/1.07	  f1#(I28, I29) -> f2#(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28]
0.99/1.07	R = 
0.99/1.07	  init(x1, x2) -> f3(rnd1, rnd2) 
0.99/1.07	  f5(I0, I1) -> f5(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.99/1.07	  f3(I2, I3) -> f5(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.99/1.07	  f4(I5, I6) -> f4(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1]
0.99/1.07	  f4(I9, I10) -> f4(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1]
0.99/1.07	  f4(I14, I15) -> f4(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1]
0.99/1.07	  f2(I19, I20) -> f4(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1]
0.99/1.07	  f3(I24, I25) -> f2(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1]
0.99/1.07	  f1(I28, I29) -> f2(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28]
0.99/1.07	
0.99/1.07	The dependency graph for this problem is:
0.99/1.07	  0 -> 2, 7
0.99/1.07	  1 -> 1
0.99/1.07	  2 -> 1
0.99/1.07	  3 -> 3, 4, 5
0.99/1.07	  4 -> 3, 4, 5
0.99/1.07	  5 -> 3, 4, 5
0.99/1.07	  6 -> 3, 4, 5
0.99/1.07	  7 -> 6
0.99/1.07	  8 -> 6
0.99/1.07	Where:
0.99/1.07	  0) init#(x1, x2) -> f3#(rnd1, rnd2) 
0.99/1.07	  1) f5#(I0, I1) -> f5#(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.99/1.07	  2) f3#(I2, I3) -> f5#(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.99/1.07	  3) f4#(I5, I6) -> f4#(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1]
0.99/1.07	  4) f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1]
0.99/1.07	  5) f4#(I14, I15) -> f4#(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1]
0.99/1.07	  6) f2#(I19, I20) -> f4#(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1]
0.99/1.07	  7) f3#(I24, I25) -> f2#(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1]
0.99/1.07	  8) f1#(I28, I29) -> f2#(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28]
0.99/1.07	
0.99/1.07	We have the following SCCs.
0.99/1.07	  { 1 }
0.99/1.07	  { 3, 4, 5 }
0.99/1.07	
0.99/1.07	DP problem for innermost termination.
0.99/1.07	P =
0.99/1.07	  f4#(I5, I6) -> f4#(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1]
0.99/1.07	  f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1]
0.99/1.07	  f4#(I14, I15) -> f4#(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1]
0.99/1.07	R = 
0.99/1.07	  init(x1, x2) -> f3(rnd1, rnd2) 
0.99/1.07	  f5(I0, I1) -> f5(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.99/1.07	  f3(I2, I3) -> f5(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.99/1.07	  f4(I5, I6) -> f4(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1]
0.99/1.07	  f4(I9, I10) -> f4(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1]
0.99/1.07	  f4(I14, I15) -> f4(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1]
0.99/1.07	  f2(I19, I20) -> f4(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1]
0.99/1.07	  f3(I24, I25) -> f2(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1]
0.99/1.07	  f1(I28, I29) -> f2(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28]
0.99/1.07	
0.99/1.07	We use the basic value criterion with the projection function NU:
0.99/1.07	  NU[f4#(z1,z2)] = z1
0.99/1.07	
0.99/1.07	This gives the following inequalities:
0.99/1.07	  I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1  ==>  I5 >! I7
0.99/1.07	  I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1  ==>  I9 (>! \union =) I11
0.99/1.07	  I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1  ==>  I14 >! I16
0.99/1.07	
0.99/1.07	We remove all the strictly oriented dependency pairs.
0.99/1.07	
0.99/1.07	DP problem for innermost termination.
0.99/1.07	P =
0.99/1.07	  f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1]
0.99/1.07	R = 
0.99/1.07	  init(x1, x2) -> f3(rnd1, rnd2) 
0.99/1.07	  f5(I0, I1) -> f5(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.99/1.07	  f3(I2, I3) -> f5(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.99/1.07	  f4(I5, I6) -> f4(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1]
0.99/1.07	  f4(I9, I10) -> f4(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1]
0.99/1.07	  f4(I14, I15) -> f4(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1]
0.99/1.07	  f2(I19, I20) -> f4(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1]
0.99/1.07	  f3(I24, I25) -> f2(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1]
0.99/1.07	  f1(I28, I29) -> f2(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28]
0.99/1.07	
0.99/4.05	EOF