8.89/8.81	YES
8.89/8.81	
8.89/8.81	DP problem for innermost termination.
8.89/8.81	P =
8.89/8.81	  f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 
8.89/8.81	  f6#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
8.89/8.81	  f6#(I6, I7, I8, I9, I10, I11) -> f5#(I6, I7, I8, I9, I10, I11) 
8.89/8.81	  f6#(I12, I13, I14, I15, I16, I17) -> f4#(I12, I13, I14, I15, I16, I17) 
8.89/8.81	  f6#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
8.89/8.81	  f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) 
8.89/8.81	  f5#(I36, I37, I38, I39, I40, I41) -> f4#(I40, I41, I38, I39, I40, I41) [I40 <= 0]
8.89/8.81	  f5#(I42, I43, I44, I45, I46, I47) -> f4#(I46, I47, I44, I45, I46, I47) [I47 <= 0]
8.89/8.81	  f5#(I48, I49, I50, I51, I52, I53) -> f1#(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52]
8.89/8.81	  f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.81	R = 
8.89/8.81	  f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 
8.89/8.81	  f6(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
8.89/8.81	  f6(I6, I7, I8, I9, I10, I11) -> f5(I6, I7, I8, I9, I10, I11) 
8.89/8.81	  f6(I12, I13, I14, I15, I16, I17) -> f4(I12, I13, I14, I15, I16, I17) 
8.89/8.81	  f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
8.89/8.81	  f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) 
8.89/8.81	  f3(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, I32, I33, I34, I35) 
8.89/8.81	  f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I38, I39, I40, I41) [I40 <= 0]
8.89/8.81	  f5(I42, I43, I44, I45, I46, I47) -> f4(I46, I47, I44, I45, I46, I47) [I47 <= 0]
8.89/8.81	  f5(I48, I49, I50, I51, I52, I53) -> f1(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52]
8.89/8.81	  f4(I54, I55, I56, I57, I58, I59) -> f2(I58, I59, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ 1 + I58 <= 0 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
8.89/8.81	  f4(I60, I61, I62, I63, I64, I65) -> f2(I64, I65, I66, I67, I68, I69) [I69 = I67 /\ I68 = I66 /\ 1 <= I64 /\ I67 = I67 /\ I66 = I66]
8.89/8.81	  f4(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ 1 + I75 <= 0 /\ I77 = I77 /\ I76 = I76]
8.89/8.81	  f4(I80, I81, I82, I83, I84, I85) -> f2(I84, I85, I86, I87, I88, I89) [I89 = I87 /\ I88 = I86 /\ 1 <= I85 /\ I87 = I87 /\ I86 = I86]
8.89/8.81	  f4(I90, I91, I92, I93, I94, I95) -> f2(I94, I95, I96, I97, I98, I99) [I99 = I97 /\ I98 = I96 /\ 0 <= I95 /\ I95 <= 0 /\ 0 <= I94 /\ I94 <= 0 /\ I97 = I97 /\ I96 = I96]
8.89/8.81	  f1(I100, I101, I102, I103, I104, I105) -> f3(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.81	  f1(I106, I107, I108, I109, I110, I111) -> f2(I110, I111, I112, I113, I114, I115) [I115 = I113 /\ I114 = I112 /\ I113 = I113 /\ I112 = I112]
8.89/8.81	
8.89/8.81	The dependency graph for this problem is:
8.89/8.81	  0 -> 1, 2, 3, 4
8.89/8.81	  1 -> 5
8.89/8.81	  2 -> 6, 7, 8
8.89/8.81	  3 -> 
8.89/8.81	  4 -> 9
8.89/8.81	  5 -> 6, 7, 8
8.89/8.81	  6 -> 
8.89/8.81	  7 -> 
8.89/8.81	  8 -> 9
8.89/8.81	  9 -> 5
8.89/8.81	Where:
8.89/8.81	  0) f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 
8.89/8.81	  1) f6#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 
8.89/8.81	  2) f6#(I6, I7, I8, I9, I10, I11) -> f5#(I6, I7, I8, I9, I10, I11) 
8.89/8.81	  3) f6#(I12, I13, I14, I15, I16, I17) -> f4#(I12, I13, I14, I15, I16, I17) 
8.89/8.81	  4) f6#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
8.89/8.81	  5) f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) 
8.89/8.81	  6) f5#(I36, I37, I38, I39, I40, I41) -> f4#(I40, I41, I38, I39, I40, I41) [I40 <= 0]
8.89/8.81	  7) f5#(I42, I43, I44, I45, I46, I47) -> f4#(I46, I47, I44, I45, I46, I47) [I47 <= 0]
8.89/8.81	  8) f5#(I48, I49, I50, I51, I52, I53) -> f1#(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52]
8.89/8.81	  9) f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.81	
8.89/8.81	We have the following SCCs.
8.89/8.81	  { 5, 8, 9 }
8.89/8.81	
8.89/8.81	DP problem for innermost termination.
8.89/8.81	P =
8.89/8.81	  f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) 
8.89/8.81	  f5#(I48, I49, I50, I51, I52, I53) -> f1#(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52]
8.89/8.81	  f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.81	R = 
8.89/8.81	  f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 
8.89/8.81	  f6(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
8.89/8.81	  f6(I6, I7, I8, I9, I10, I11) -> f5(I6, I7, I8, I9, I10, I11) 
8.89/8.81	  f6(I12, I13, I14, I15, I16, I17) -> f4(I12, I13, I14, I15, I16, I17) 
8.89/8.81	  f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
8.89/8.82	  f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) 
8.89/8.82	  f3(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, I32, I33, I34, I35) 
8.89/8.82	  f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I38, I39, I40, I41) [I40 <= 0]
8.89/8.82	  f5(I42, I43, I44, I45, I46, I47) -> f4(I46, I47, I44, I45, I46, I47) [I47 <= 0]
8.89/8.82	  f5(I48, I49, I50, I51, I52, I53) -> f1(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52]
8.89/8.82	  f4(I54, I55, I56, I57, I58, I59) -> f2(I58, I59, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ 1 + I58 <= 0 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
8.89/8.82	  f4(I60, I61, I62, I63, I64, I65) -> f2(I64, I65, I66, I67, I68, I69) [I69 = I67 /\ I68 = I66 /\ 1 <= I64 /\ I67 = I67 /\ I66 = I66]
8.89/8.82	  f4(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ 1 + I75 <= 0 /\ I77 = I77 /\ I76 = I76]
8.89/8.82	  f4(I80, I81, I82, I83, I84, I85) -> f2(I84, I85, I86, I87, I88, I89) [I89 = I87 /\ I88 = I86 /\ 1 <= I85 /\ I87 = I87 /\ I86 = I86]
8.89/8.82	  f4(I90, I91, I92, I93, I94, I95) -> f2(I94, I95, I96, I97, I98, I99) [I99 = I97 /\ I98 = I96 /\ 0 <= I95 /\ I95 <= 0 /\ 0 <= I94 /\ I94 <= 0 /\ I97 = I97 /\ I96 = I96]
8.89/8.82	  f1(I100, I101, I102, I103, I104, I105) -> f3(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.82	  f1(I106, I107, I108, I109, I110, I111) -> f2(I110, I111, I112, I113, I114, I115) [I115 = I113 /\ I114 = I112 /\ I113 = I113 /\ I112 = I112]
8.89/8.82	
8.89/8.82	We use the extended value criterion with the projection function NU:
8.89/8.82	  NU[f1#(x0,x1,x2,x3,x4,x5)] = x5
8.89/8.82	  NU[f5#(x0,x1,x2,x3,x4,x5)] = x5 + 1
8.89/8.82	  NU[f3#(x0,x1,x2,x3,x4,x5)] = x5 + 1
8.89/8.82	
8.89/8.82	This gives the following inequalities:
8.89/8.82	    ==>  I35 + 1 >= I35 + 1
8.89/8.82	  1 <= I53 /\ 1 <= I52  ==>  I53 + 1 > I53  with  I53 + 1 >= 0
8.89/8.82	    ==>  I105 >= (-1 + I105) + 1
8.89/8.82	
8.89/8.82	We remove all the strictly oriented dependency pairs.
8.89/8.82	
8.89/8.82	DP problem for innermost termination.
8.89/8.82	P =
8.89/8.82	  f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) 
8.89/8.82	  f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.82	R = 
8.89/8.82	  f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 
8.89/8.82	  f6(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 
8.89/8.82	  f6(I6, I7, I8, I9, I10, I11) -> f5(I6, I7, I8, I9, I10, I11) 
8.89/8.82	  f6(I12, I13, I14, I15, I16, I17) -> f4(I12, I13, I14, I15, I16, I17) 
8.89/8.82	  f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
8.89/8.82	  f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) 
8.89/8.82	  f3(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, I32, I33, I34, I35) 
8.89/8.82	  f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I38, I39, I40, I41) [I40 <= 0]
8.89/8.82	  f5(I42, I43, I44, I45, I46, I47) -> f4(I46, I47, I44, I45, I46, I47) [I47 <= 0]
8.89/8.82	  f5(I48, I49, I50, I51, I52, I53) -> f1(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52]
8.89/8.82	  f4(I54, I55, I56, I57, I58, I59) -> f2(I58, I59, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ 1 + I58 <= 0 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
8.89/8.82	  f4(I60, I61, I62, I63, I64, I65) -> f2(I64, I65, I66, I67, I68, I69) [I69 = I67 /\ I68 = I66 /\ 1 <= I64 /\ I67 = I67 /\ I66 = I66]
8.89/8.82	  f4(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ 1 + I75 <= 0 /\ I77 = I77 /\ I76 = I76]
8.89/8.82	  f4(I80, I81, I82, I83, I84, I85) -> f2(I84, I85, I86, I87, I88, I89) [I89 = I87 /\ I88 = I86 /\ 1 <= I85 /\ I87 = I87 /\ I86 = I86]
8.89/8.82	  f4(I90, I91, I92, I93, I94, I95) -> f2(I94, I95, I96, I97, I98, I99) [I99 = I97 /\ I98 = I96 /\ 0 <= I95 /\ I95 <= 0 /\ 0 <= I94 /\ I94 <= 0 /\ I97 = I97 /\ I96 = I96]
8.89/8.82	  f1(I100, I101, I102, I103, I104, I105) -> f3(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.82	  f1(I106, I107, I108, I109, I110, I111) -> f2(I110, I111, I112, I113, I114, I115) [I115 = I113 /\ I114 = I112 /\ I113 = I113 /\ I112 = I112]
8.89/8.82	
8.89/8.82	The dependency graph for this problem is:
8.89/8.82	  5 -> 
8.89/8.82	  9 -> 5
8.89/8.82	Where:
8.89/8.82	  5) f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) 
8.89/8.82	  9) f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) 
8.89/8.82	
8.89/8.82	We have the following SCCs.
8.89/8.82	  
8.89/11.79	EOF