25.96/25.65	YES
25.96/25.65	
25.96/25.65	DP problem for innermost termination.
25.96/25.65	P =
25.96/25.65	  f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
25.96/25.65	  f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	  f4#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
25.96/25.65	  f2#(I28, I29, I30, I31, I32, I33, I34) -> f3#(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
25.96/25.65	  f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2]
25.96/25.65	R = 
25.96/25.65	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
25.96/25.65	  f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	  f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
25.96/25.65	  f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
25.96/25.65	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2]
25.96/25.65	
25.96/25.65	The dependency graph for this problem is:
25.96/25.65	  0 -> 6
25.96/25.65	  1 -> 2, 3
25.96/25.65	  2 -> 1
25.96/25.65	  3 -> 5
25.96/25.65	  4 -> 2, 3
25.96/25.65	  5 -> 2
25.96/25.65	  6 -> 5
25.96/25.65	Where:
25.96/25.65	  0) f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
25.96/25.65	  1) f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  2) f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  3) f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	  4) f4#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
25.96/25.65	  5) f2#(I28, I29, I30, I31, I32, I33, I34) -> f3#(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
25.96/25.65	  6) f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2]
25.96/25.65	
25.96/25.65	We have the following SCCs.
25.96/25.65	  { 1, 2, 3, 5 }
25.96/25.65	
25.96/25.65	DP problem for innermost termination.
25.96/25.65	P =
25.96/25.65	  f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	  f2#(I28, I29, I30, I31, I32, I33, I34) -> f3#(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
25.96/25.65	R = 
25.96/25.65	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
25.96/25.65	  f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	  f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
25.96/25.65	  f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
25.96/25.65	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2]
25.96/25.65	
25.96/25.65	We use the extended value criterion with the projection function NU:
25.96/25.65	  NU[f2#(x0,x1,x2,x3,x4,x5,x6)] = x1 - 1
25.96/25.65	  NU[f3#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x4 - 1
25.96/25.65	  NU[f5#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x4 - 1
25.96/25.65	
25.96/25.65	This gives the following inequalities:
25.96/25.65	    ==>  I1 - I4 - 1 >= I1 - I4 - 1
25.96/25.65	  1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4  ==>  I8 - I11 - 1 >= (-1 + I8) - (-1 + I11) - 1
25.96/25.65	  I18 <= 0 /\ I18 <= 0  ==>  I15 - I18 - 1 >= I15 - 1
25.96/25.65	  1 <= I29 /\ 1 <= I28 /\ 1 <= I29  ==>  I29 - 1 > I29 - I28 - 1  with  I29 - 1 >= 0
25.96/25.65	
25.96/25.65	We remove all the strictly oriented dependency pairs.
25.96/25.65	
25.96/25.65	DP problem for innermost termination.
25.96/25.65	P =
25.96/25.65	  f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	R = 
25.96/25.65	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
25.96/25.65	  f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	  f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
25.96/25.65	  f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
25.96/25.65	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2]
25.96/25.65	
25.96/25.65	The dependency graph for this problem is:
25.96/25.65	  1 -> 2, 3
25.96/25.65	  2 -> 1
25.96/25.65	  3 -> 
25.96/25.65	Where:
25.96/25.65	  1) f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  2) f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  3) f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	
25.96/25.65	We have the following SCCs.
25.96/25.65	  { 1, 2 }
25.96/25.65	
25.96/25.65	DP problem for innermost termination.
25.96/25.65	P =
25.96/25.65	  f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	R = 
25.96/25.65	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
25.96/25.65	  f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 
25.96/25.65	  f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
25.96/25.65	  f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
25.96/25.65	  f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
25.96/25.65	  f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
25.96/25.65	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2]
25.96/25.65	
25.96/25.65	We use the basic value criterion with the projection function NU:
25.96/25.65	  NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z5
25.96/25.65	  NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z5
25.96/25.65	
25.96/25.65	This gives the following inequalities:
25.96/25.65	    ==>  I4 (>! \union =) I4
25.96/25.65	  1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4  ==>  I11 >! -1 + I11
25.96/25.65	
25.96/25.65	We remove all the strictly oriented dependency pairs.
25.96/25.65	
25.96/25.65	DP problem for innermost termination.
25.96/25.65	P =
25.96/25.65	  f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
26.04/25.65	R = 
26.04/25.65	  f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
26.04/25.65	  f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 
26.04/25.65	  f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
26.04/25.65	  f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0]
26.04/25.65	  f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3]
26.04/25.65	  f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29]
26.04/25.65	  f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2]
26.04/25.65	
26.04/25.65	The dependency graph for this problem is:
26.04/25.65	  1 -> 
26.04/25.65	Where:
26.04/25.65	  1) f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 
26.04/25.65	
26.04/25.65	We have the following SCCs.
26.04/25.65	  
26.04/28.63	EOF