12.16/12.06	YES
12.16/12.06	
12.16/12.06	DP problem for innermost termination.
12.16/12.06	P =
12.16/12.06	  f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) 
12.16/12.06	  f8#(I0, I1, I2, I3, I4, I5) -> f7#(I0, I1, I2, I3, I4, I5) 
12.16/12.06	  f8#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, I8, I9, I10, I11) 
12.16/12.06	  f8#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, I16, I17) 
12.16/12.06	  f8#(I18, I19, I20, I21, I22, I23) -> f4#(I18, I19, I20, I21, I22, I23) 
12.16/12.06	  f8#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, I28, I29) 
12.16/12.06	  f8#(I30, I31, I32, I33, I34, I35) -> f1#(I30, I31, I32, I33, I34, I35) 
12.16/12.06	  f8#(I42, I43, I44, I45, I46, I47) -> f7#(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
12.16/12.06	  f7#(I48, I49, I50, I51, I52, I53) -> f6#(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54]
12.16/12.06	  f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  f6#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, I68, 5) [100 <= I68]
12.16/12.06	  f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	  f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  f4#(I84, I85, I86, I87, I88, I89) -> f1#(I88, I89, I86, I87, I88, I89) [21 <= I89]
12.16/12.06	  f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	R = 
12.16/12.06	  f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 
12.16/12.06	  f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) 
12.16/12.06	  f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) 
12.16/12.06	  f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) 
12.16/12.06	  f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) 
12.16/12.06	  f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) 
12.16/12.06	  f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) 
12.16/12.06	  f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 
12.16/12.06	  f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
12.16/12.06	  f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54]
12.16/12.06	  f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68]
12.16/12.06	  f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	  f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89]
12.16/12.06	  f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	  f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102]
12.16/12.06	
12.16/12.06	The dependency graph for this problem is:
12.16/12.06	  0 -> 1, 2, 3, 4, 5, 6, 7
12.16/12.06	  1 -> 8
12.16/12.06	  2 -> 9, 10
12.16/12.06	  3 -> 11
12.16/12.06	  4 -> 12, 13
12.16/12.06	  5 -> 14
12.16/12.06	  6 -> 
12.16/12.06	  7 -> 8
12.16/12.06	  8 -> 9
12.16/12.06	  9 -> 11
12.16/12.06	  10 -> 12
12.16/12.06	  11 -> 9, 10
12.16/12.06	  12 -> 14
12.16/12.06	  13 -> 
12.16/12.06	  14 -> 12, 13
12.16/12.06	Where:
12.16/12.06	  0) f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) 
12.16/12.06	  1) f8#(I0, I1, I2, I3, I4, I5) -> f7#(I0, I1, I2, I3, I4, I5) 
12.16/12.06	  2) f8#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, I8, I9, I10, I11) 
12.16/12.06	  3) f8#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, I16, I17) 
12.16/12.06	  4) f8#(I18, I19, I20, I21, I22, I23) -> f4#(I18, I19, I20, I21, I22, I23) 
12.16/12.06	  5) f8#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, I28, I29) 
12.16/12.06	  6) f8#(I30, I31, I32, I33, I34, I35) -> f1#(I30, I31, I32, I33, I34, I35) 
12.16/12.06	  7) f8#(I42, I43, I44, I45, I46, I47) -> f7#(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
12.16/12.06	  8) f7#(I48, I49, I50, I51, I52, I53) -> f6#(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54]
12.16/12.06	  9) f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  10) f6#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, I68, 5) [100 <= I68]
12.16/12.06	  11) f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	  12) f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  13) f4#(I84, I85, I86, I87, I88, I89) -> f1#(I88, I89, I86, I87, I88, I89) [21 <= I89]
12.16/12.06	  14) f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	
12.16/12.06	We have the following SCCs.
12.16/12.06	  { 9, 11 }
12.16/12.06	  { 12, 14 }
12.16/12.06	
12.16/12.06	DP problem for innermost termination.
12.16/12.06	P =
12.16/12.06	  f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	R = 
12.16/12.06	  f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 
12.16/12.06	  f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) 
12.16/12.06	  f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) 
12.16/12.06	  f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) 
12.16/12.06	  f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) 
12.16/12.06	  f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) 
12.16/12.06	  f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) 
12.16/12.06	  f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 
12.16/12.06	  f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
12.16/12.06	  f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54]
12.16/12.06	  f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68]
12.16/12.06	  f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	  f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89]
12.16/12.06	  f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	  f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102]
12.16/12.06	
12.16/12.06	We use the reverse value criterion with the projection function NU:
12.16/12.06	  NU[f3#(z1,z2,z3,z4,z5,z6)] = 20 + -1 * (3 + z6)
12.16/12.06	  NU[f4#(z1,z2,z3,z4,z5,z6)] = 20 + -1 * z6
12.16/12.06	
12.16/12.06	This gives the following inequalities:
12.16/12.06	  I83 <= 20  ==>  20 + -1 * I83 > 20 + -1 * (3 + I83)  with  20 + -1 * I83 >= 0
12.16/12.06	    ==>  20 + -1 * (3 + I95) >= 20 + -1 * (3 + I95)
12.16/12.06	
12.16/12.06	We remove all the strictly oriented dependency pairs.
12.16/12.06	
12.16/12.06	DP problem for innermost termination.
12.16/12.06	P =
12.16/12.06	  f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	R = 
12.16/12.06	  f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 
12.16/12.06	  f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) 
12.16/12.06	  f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) 
12.16/12.06	  f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) 
12.16/12.06	  f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) 
12.16/12.06	  f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) 
12.16/12.06	  f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) 
12.16/12.06	  f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 
12.16/12.06	  f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
12.16/12.06	  f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54]
12.16/12.06	  f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68]
12.16/12.06	  f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	  f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89]
12.16/12.06	  f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	  f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102]
12.16/12.06	
12.16/12.06	The dependency graph for this problem is:
12.16/12.06	  14 -> 
12.16/12.06	Where:
12.16/12.06	  14) f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	
12.16/12.06	We have the following SCCs.
12.16/12.06	  
12.16/12.06	
12.16/12.06	DP problem for innermost termination.
12.16/12.06	P =
12.16/12.06	  f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	R = 
12.16/12.06	  f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 
12.16/12.06	  f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) 
12.16/12.06	  f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) 
12.16/12.06	  f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) 
12.16/12.06	  f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) 
12.16/12.06	  f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) 
12.16/12.06	  f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) 
12.16/12.06	  f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 
12.16/12.06	  f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
12.16/12.06	  f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54]
12.16/12.06	  f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68]
12.16/12.06	  f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	  f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89]
12.16/12.06	  f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	  f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102]
12.16/12.06	
12.16/12.06	We use the reverse value criterion with the projection function NU:
12.16/12.06	  NU[f5#(z1,z2,z3,z4,z5,z6)] = 99 + -1 * (1 + z5)
12.16/12.06	  NU[f6#(z1,z2,z3,z4,z5,z6)] = 99 + -1 * z5
12.16/12.06	
12.16/12.06	This gives the following inequalities:
12.16/12.06	  I63 = I62 /\ I60 <= 99 /\ I62 = I62  ==>  99 + -1 * I60 > 99 + -1 * (1 + I60)  with  99 + -1 * I60 >= 0
12.16/12.06	  I77 = I76 /\ I76 = I76  ==>  99 + -1 * (1 + I74) >= 99 + -1 * (1 + I74)
12.16/12.06	
12.16/12.06	We remove all the strictly oriented dependency pairs.
12.16/12.06	
12.16/12.06	DP problem for innermost termination.
12.16/12.06	P =
12.16/12.06	  f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	R = 
12.16/12.06	  f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 
12.16/12.06	  f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) 
12.16/12.06	  f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) 
12.16/12.06	  f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) 
12.16/12.06	  f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) 
12.16/12.06	  f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) 
12.16/12.06	  f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) 
12.16/12.06	  f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 
12.16/12.06	  f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
12.16/12.06	  f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54]
12.16/12.06	  f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62]
12.16/12.06	  f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68]
12.16/12.06	  f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	  f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20]
12.16/12.06	  f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89]
12.16/12.06	  f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) 
12.16/12.06	  f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102]
12.16/12.06	
12.16/12.06	The dependency graph for this problem is:
12.16/12.06	  11 -> 
12.16/12.06	Where:
12.16/12.06	  11) f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76]
12.16/12.06	
12.16/12.06	We have the following SCCs.
12.16/12.06	  
12.16/15.03	EOF