5.21/5.21	YES
5.21/5.21	
5.21/5.21	DP problem for innermost termination.
5.21/5.21	P =
5.21/5.21	  init#(x1, x2, x3, x4, x5, x6) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5, rnd6) 
5.21/5.21	  f3#(I0, I1, I2, I3, I4, I5) -> f3#(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  f3#(I8, I9, I10, I11, I12, I13) -> f2#(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	  f2#(I20, I21, I22, I23, I24, I25) -> f3#(I26, 1, 1, 1, I27, I28) [I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20]
5.21/5.21	  f1#(I29, I30, I31, I32, I33, I34) -> f2#(I35, I36, I37, I38, I39, I40) [3 <= I35 - 1]
5.21/5.21	R = 
5.21/5.21	  init(x1, x2, x3, x4, x5, x6) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5, rnd6) 
5.21/5.21	  f3(I0, I1, I2, I3, I4, I5) -> f3(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  f3(I8, I9, I10, I11, I12, I13) -> f2(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	  f2(I20, I21, I22, I23, I24, I25) -> f3(I26, 1, 1, 1, I27, I28) [I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20]
5.21/5.21	  f1(I29, I30, I31, I32, I33, I34) -> f2(I35, I36, I37, I38, I39, I40) [3 <= I35 - 1]
5.21/5.21	
5.21/5.21	The dependency graph for this problem is:
5.21/5.21	  0 -> 4
5.21/5.21	  1 -> 1, 2
5.21/5.21	  2 -> 3
5.21/5.21	  3 -> 1, 2
5.21/5.21	  4 -> 3
5.21/5.21	Where:
5.21/5.21	  0) init#(x1, x2, x3, x4, x5, x6) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5, rnd6) 
5.21/5.21	  1) f3#(I0, I1, I2, I3, I4, I5) -> f3#(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  2) f3#(I8, I9, I10, I11, I12, I13) -> f2#(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	  3) f2#(I20, I21, I22, I23, I24, I25) -> f3#(I26, 1, 1, 1, I27, I28) [I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20]
5.21/5.21	  4) f1#(I29, I30, I31, I32, I33, I34) -> f2#(I35, I36, I37, I38, I39, I40) [3 <= I35 - 1]
5.21/5.21	
5.21/5.21	We have the following SCCs.
5.21/5.21	  { 1, 2, 3 }
5.21/5.21	
5.21/5.21	DP problem for innermost termination.
5.21/5.21	P =
5.21/5.21	  f3#(I0, I1, I2, I3, I4, I5) -> f3#(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  f3#(I8, I9, I10, I11, I12, I13) -> f2#(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	  f2#(I20, I21, I22, I23, I24, I25) -> f3#(I26, 1, 1, 1, I27, I28) [I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20]
5.21/5.21	R = 
5.21/5.21	  init(x1, x2, x3, x4, x5, x6) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5, rnd6) 
5.21/5.21	  f3(I0, I1, I2, I3, I4, I5) -> f3(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  f3(I8, I9, I10, I11, I12, I13) -> f2(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	  f2(I20, I21, I22, I23, I24, I25) -> f3(I26, 1, 1, 1, I27, I28) [I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20]
5.21/5.21	  f1(I29, I30, I31, I32, I33, I34) -> f2(I35, I36, I37, I38, I39, I40) [3 <= I35 - 1]
5.21/5.21	
5.21/5.21	We use the basic value criterion with the projection function NU:
5.21/5.21	  NU[f2#(z1,z2,z3,z4,z5,z6)] = z1
5.21/5.21	  NU[f3#(z1,z2,z3,z4,z5,z6)] = z6
5.21/5.21	
5.21/5.21	This gives the following inequalities:
5.21/5.21	  I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4  ==>  I5 (>! \union =) I7
5.21/5.21	  I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1  ==>  I13 (>! \union =) I14
5.21/5.21	  I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20  ==>  I20 >! I28
5.21/5.21	
5.21/5.21	We remove all the strictly oriented dependency pairs.
5.21/5.21	
5.21/5.21	DP problem for innermost termination.
5.21/5.21	P =
5.21/5.21	  f3#(I0, I1, I2, I3, I4, I5) -> f3#(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  f3#(I8, I9, I10, I11, I12, I13) -> f2#(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	R = 
5.21/5.21	  init(x1, x2, x3, x4, x5, x6) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5, rnd6) 
5.21/5.21	  f3(I0, I1, I2, I3, I4, I5) -> f3(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  f3(I8, I9, I10, I11, I12, I13) -> f2(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	  f2(I20, I21, I22, I23, I24, I25) -> f3(I26, 1, 1, 1, I27, I28) [I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20]
5.21/5.21	  f1(I29, I30, I31, I32, I33, I34) -> f2(I35, I36, I37, I38, I39, I40) [3 <= I35 - 1]
5.21/5.21	
5.21/5.21	The dependency graph for this problem is:
5.21/5.21	  1 -> 1, 2
5.21/5.21	  2 -> 
5.21/5.21	Where:
5.21/5.21	  1) f3#(I0, I1, I2, I3, I4, I5) -> f3#(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  2) f3#(I8, I9, I10, I11, I12, I13) -> f2#(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	
5.21/5.21	We have the following SCCs.
5.21/5.21	  { 1 }
5.21/5.21	
5.21/5.21	DP problem for innermost termination.
5.21/5.21	P =
5.21/5.21	  f3#(I0, I1, I2, I3, I4, I5) -> f3#(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	R = 
5.21/5.21	  init(x1, x2, x3, x4, x5, x6) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5, rnd6) 
5.21/5.21	  f3(I0, I1, I2, I3, I4, I5) -> f3(I6, I1 * I2, I2 + 1, I2 + 1, I4, I7) [I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4]
5.21/5.21	  f3(I8, I9, I10, I11, I12, I13) -> f2(I14, I15, I16, I17, I18, I19) [I10 = I11 /\ I12 + 2 <= I8 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ 0 <= I8 - 1 /\ I14 <= I13 /\ I14 + 1 <= I8 /\ I12 <= I10 - 1 /\ 0 <= I9 - 1]
5.21/5.21	  f2(I20, I21, I22, I23, I24, I25) -> f3(I26, 1, 1, 1, I27, I28) [I27 + 2 <= I20 /\ -1 <= I28 - 1 /\ 0 <= I26 - 1 /\ 0 <= I20 - 1 /\ I28 + 1 <= I20 /\ I26 <= I20]
5.21/5.21	  f1(I29, I30, I31, I32, I33, I34) -> f2(I35, I36, I37, I38, I39, I40) [3 <= I35 - 1]
5.21/5.21	
5.21/5.21	We use the reverse value criterion with the projection function NU:
5.21/5.21	  NU[f3#(z1,z2,z3,z4,z5,z6)] = z5 + -1 * z3
5.21/5.21	
5.21/5.21	This gives the following inequalities:
5.21/5.21	  I2 = I3 /\ I4 + 2 <= I0 /\ -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ I7 <= I5 /\ I7 + 1 <= I0 /\ I6 <= I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ I2 <= I4  ==>  I4 + -1 * I2 > I4 + -1 * (I2 + 1)  with  I4 + -1 * I2 >= 0
5.21/5.21	
5.21/5.21	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
5.21/8.19	EOF