6.85/6.81	YES
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
6.85/6.81	  f10#(I0, I1, I2, I3, I4) -> f4#(I0, 1, I2, I3, I4) 
6.85/6.81	  f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I5, I9) 
6.85/6.81	  f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	  f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 
6.85/6.81	  f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 
6.85/6.81	  f4#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4#(I50, I51, I52, I53, I54) -> f2#(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	The dependency graph for this problem is:
6.85/6.81	  0 -> 1
6.85/6.81	  1 -> 11
6.85/6.81	  2 -> 7
6.85/6.81	  3 -> 7
6.85/6.81	  4 -> 5
6.85/6.81	  5 -> 6
6.85/6.81	  6 -> 5
6.85/6.81	  7 -> 3, 4
6.85/6.81	  8 -> 12, 13
6.85/6.81	  9 -> 8
6.85/6.81	  10 -> 2
6.85/6.81	  11 -> 2
6.85/6.81	  12 -> 8
6.85/6.81	  13 -> 2
6.85/6.81	Where:
6.85/6.81	  0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
6.85/6.81	  1) f10#(I0, I1, I2, I3, I4) -> f4#(I0, 1, I2, I3, I4) 
6.85/6.81	  2) f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I5, I9) 
6.85/6.81	  3) f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  4) f6#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	  6) f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  7) f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 
6.85/6.81	  8) f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 
6.85/6.81	  9) f4#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  10) f4#(I50, I51, I52, I53, I54) -> f2#(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  11) f4#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  12) f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  13) f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	We have the following SCCs.
6.85/6.81	  { 8, 12 }
6.85/6.81	  { 3, 7 }
6.85/6.81	  { 5, 6 }
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	  f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	We use the basic value criterion with the projection function NU:
6.85/6.81	  NU[f7#(z1,z2,z3,z4,z5)] = z5
6.85/6.81	  NU[f9#(z1,z2,z3,z4,z5)] = z5
6.85/6.81	
6.85/6.81	This gives the following inequalities:
6.85/6.81	    ==>  I24 (>! \union =) I24
6.85/6.81	  0 <= I29  ==>  I29 >! -1 + I29
6.85/6.81	
6.85/6.81	We remove all the strictly oriented dependency pairs.
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	The dependency graph for this problem is:
6.85/6.81	  5 -> 
6.85/6.81	Where:
6.85/6.81	  5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	
6.85/6.81	We have the following SCCs.
6.85/6.81	  
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	We use the basic value criterion with the projection function NU:
6.85/6.81	  NU[f5#(z1,z2,z3,z4,z5)] = z4
6.85/6.81	  NU[f6#(z1,z2,z3,z4,z5)] = z4
6.85/6.81	
6.85/6.81	This gives the following inequalities:
6.85/6.81	  0 <= I13  ==>  I13 >! -1 + I13
6.85/6.81	    ==>  I38 (>! \union =) I38
6.85/6.81	
6.85/6.81	We remove all the strictly oriented dependency pairs.
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	The dependency graph for this problem is:
6.85/6.81	  7 -> 
6.85/6.81	Where:
6.85/6.81	  7) f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 
6.85/6.81	
6.85/6.81	We have the following SCCs.
6.85/6.81	  
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 
6.85/6.81	  f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	We use the basic value criterion with the projection function NU:
6.85/6.81	  NU[f1#(z1,z2,z3,z4,z5)] = z3
6.85/6.81	  NU[f3#(z1,z2,z3,z4,z5)] = z3
6.85/6.81	
6.85/6.81	This gives the following inequalities:
6.85/6.81	    ==>  I42 (>! \union =) I42
6.85/6.81	  0 <= I62  ==>  I62 >! -1 + I62
6.85/6.81	
6.85/6.81	We remove all the strictly oriented dependency pairs.
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	The dependency graph for this problem is:
6.85/6.81	  8 -> 
6.85/6.81	Where:
6.85/6.81	  8) f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 
6.85/6.81	
6.85/6.81	We have the following SCCs.
6.85/6.81	  
6.85/9.79	EOF