24.58/24.29	YES
24.58/24.29	
24.58/24.29	DP problem for innermost termination.
24.58/24.29	P =
24.58/24.29	  f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8#(I0, I1, I2, I3, I4, I5, I6) -> f1#(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7#(I28, I29, I30, I31, I32, I33, I34) -> f6#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7#(I35, I36, I37, I38, I39, I40, I41) -> f5#(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	R = 
24.58/24.29	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57]
24.58/24.29	  f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	The dependency graph for this problem is:
24.58/24.29	  0 -> 1
24.58/24.29	  1 -> 9
24.58/24.29	  2 -> 9
24.58/24.29	  3 -> 7
24.58/24.29	  4 -> 8
24.58/24.29	  5 -> 7
24.58/24.29	  6 -> 4
24.58/24.29	  7 -> 5, 6
24.58/24.29	  8 -> 4
24.58/24.29	  9 -> 2, 3
24.58/24.29	Where:
24.58/24.29	  0) f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  1) f8#(I0, I1, I2, I3, I4, I5, I6) -> f1#(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  3) f2#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  5) f7#(I28, I29, I30, I31, I32, I33, I34) -> f6#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  6) f7#(I35, I36, I37, I38, I39, I40, I41) -> f5#(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  7) f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  8) f3#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  9) f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	We have the following SCCs.
24.58/24.29	  { 2, 9 }
24.58/24.29	  { 5, 7 }
24.58/24.29	  { 4, 8 }
24.58/24.29	
24.58/24.29	DP problem for innermost termination.
24.58/24.29	P =
24.58/24.29	  f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f3#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	R = 
24.58/24.29	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57]
24.58/24.29	  f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	We use the reverse value criterion with the projection function NU:
24.58/24.29	  NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2)
24.58/24.29	  NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2)
24.58/24.29	
24.58/24.29	This gives the following inequalities:
24.58/24.29	    ==>  I24 + -1 * (1 + I22) >= I24 + -1 * (1 + I22)
24.58/24.29	  1 + I50 <= I52  ==>  I52 + -1 * (1 + I50) > I52 + -1 * (1 + (1 + I50))  with  I52 + -1 * (1 + I50) >= 0
24.58/24.29	
24.58/24.29	We remove all the strictly oriented dependency pairs.
24.58/24.29	
24.58/24.29	DP problem for innermost termination.
24.58/24.29	P =
24.58/24.29	  f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	R = 
24.58/24.29	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57]
24.58/24.29	  f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	The dependency graph for this problem is:
24.58/24.29	  4 -> 
24.58/24.29	Where:
24.58/24.29	  4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	
24.58/24.29	We have the following SCCs.
24.58/24.29	  
24.58/24.29	
24.58/24.29	DP problem for innermost termination.
24.58/24.29	P =
24.58/24.29	  f7#(I28, I29, I30, I31, I32, I33, I34) -> f6#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	R = 
24.58/24.29	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57]
24.58/24.29	  f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	We use the reverse value criterion with the projection function NU:
24.58/24.29	  NU[f6#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2)
24.58/24.29	  NU[f7#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2)
24.58/24.29	
24.58/24.29	This gives the following inequalities:
24.58/24.29	  1 + I29 <= I31  ==>  I31 + -1 * (1 + I29) > I31 + -1 * (1 + (1 + I29))  with  I31 + -1 * (1 + I29) >= 0
24.58/24.29	    ==>  I45 + -1 * (1 + I43) >= I45 + -1 * (1 + I43)
24.58/24.29	
24.58/24.29	We remove all the strictly oriented dependency pairs.
24.58/24.29	
24.58/24.29	DP problem for innermost termination.
24.58/24.29	P =
24.58/24.29	  f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	R = 
24.58/24.29	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57]
24.58/24.29	  f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	The dependency graph for this problem is:
24.58/24.29	  7 -> 
24.58/24.29	Where:
24.58/24.29	  7) f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	
24.58/24.29	We have the following SCCs.
24.58/24.29	  
24.58/24.29	
24.58/24.29	DP problem for innermost termination.
24.58/24.29	P =
24.58/24.29	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	R = 
24.58/24.29	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57]
24.58/24.29	  f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	We use the reverse value criterion with the projection function NU:
24.58/24.29	  NU[f1#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2)
24.58/24.29	  NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2)
24.58/24.29	
24.58/24.29	This gives the following inequalities:
24.58/24.29	  1 + I8 <= I10  ==>  I10 + -1 * (1 + I8) > I10 + -1 * (1 + (1 + I8))  with  I10 + -1 * (1 + I8) >= 0
24.58/24.29	    ==>  I66 + -1 * (1 + I64) >= I66 + -1 * (1 + I64)
24.58/24.29	
24.58/24.29	We remove all the strictly oriented dependency pairs.
24.58/24.29	
24.58/24.29	DP problem for innermost termination.
24.58/24.29	P =
24.58/24.29	  f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	R = 
24.58/24.29	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
24.58/24.29	  f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7]
24.58/24.29	  f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10]
24.58/24.29	  f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15]
24.58/24.29	  f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 
24.58/24.29	  f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31]
24.58/24.29	  f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36]
24.58/24.29	  f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) 
24.58/24.29	  f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52]
24.58/24.29	  f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57]
24.58/24.29	  f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	The dependency graph for this problem is:
24.58/24.29	  9 -> 
24.58/24.29	Where:
24.58/24.29	  9) f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) 
24.58/24.29	
24.58/24.29	We have the following SCCs.
24.58/24.29	  
24.58/27.27	EOF