33.88/33.45	YES
33.88/33.45	
33.88/33.45	DP problem for innermost termination.
33.88/33.45	P =
33.88/33.45	  f8#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
33.88/33.45	  f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.45	  f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.45	  f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.45	  f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.45	  f2#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.45	  f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.45	  f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.45	  f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I61, I62, I63, I64, I65, I66, I67) 
33.88/33.45	R = 
33.88/33.45	  f8(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
33.88/33.45	  f7(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.45	  f6(I7, I8, I9, I10, I11, I12, I13) -> f7(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.45	  f5(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.45	  f5(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.45	  f2(I28, I29, I30, I31, I32, I33, I34) -> f5(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.45	  f4(I37, I38, I39, I40, I41, I42, I43) -> f2(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.45	  f2(I44, I45, I46, I47, I48, I49, I50) -> f4(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.45	  f2(I54, I55, I56, I57, I58, I59, I60) -> f3(I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
33.88/33.45	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I61, I62, I63, I64, I65, I66, I67) 
33.88/33.45	
33.88/33.45	The dependency graph for this problem is:
33.88/33.45	  0 -> 8
33.88/33.45	  1 -> 5, 7
33.88/33.45	  2 -> 1
33.88/33.45	  3 -> 2
33.88/33.45	  4 -> 2
33.88/33.45	  5 -> 3, 4
33.88/33.45	  6 -> 5, 7
33.88/33.45	  7 -> 6
33.88/33.45	  8 -> 5, 7
33.88/33.45	Where:
33.88/33.45	  0) f8#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
33.88/33.45	  1) f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.45	  2) f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.45	  3) f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.45	  4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.45	  5) f2#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.45	  6) f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.45	  7) f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.45	  8) f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I61, I62, I63, I64, I65, I66, I67) 
33.88/33.45	
33.88/33.45	We have the following SCCs.
33.88/33.45	  { 1, 2, 3, 4, 5, 6, 7 }
33.88/33.45	
33.88/33.45	DP problem for innermost termination.
33.88/33.45	P =
33.88/33.45	  f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.45	  f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.45	  f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.45	  f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.45	  f2#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.45	  f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.45	  f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.45	R = 
33.88/33.45	  f8(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
33.88/33.45	  f7(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.45	  f6(I7, I8, I9, I10, I11, I12, I13) -> f7(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.45	  f5(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.45	  f5(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.45	  f2(I28, I29, I30, I31, I32, I33, I34) -> f5(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.45	  f4(I37, I38, I39, I40, I41, I42, I43) -> f2(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.45	  f2(I44, I45, I46, I47, I48, I49, I50) -> f4(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.45	  f2(I54, I55, I56, I57, I58, I59, I60) -> f3(I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
33.88/33.45	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I61, I62, I63, I64, I65, I66, I67) 
33.88/33.45	
33.88/33.45	We use the extended value criterion with the projection function NU:
33.88/33.45	  NU[f4#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 1
33.88/33.45	  NU[f5#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 2
33.88/33.45	  NU[f6#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 2
33.88/33.45	  NU[f2#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 1
33.88/33.45	  NU[f7#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 1
33.88/33.45	
33.88/33.45	This gives the following inequalities:
33.88/33.45	    ==>  I5 - 1 >= I5 - 1
33.88/33.45	  y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1  ==>  I12 - 2 >= (-1 + I12) - 1
33.88/33.45	  1 + I16 <= 0  ==>  I19 - 2 >= I19 - 2
33.88/33.45	  1 <= I23  ==>  I26 - 2 >= I26 - 2
33.88/33.45	  1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35  ==>  I33 - 1 > I33 - 2  with  I33 - 1 >= 0
33.88/33.45	    ==>  I42 - 1 >= I42 - 1
33.88/33.45	  1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50  ==>  I49 - 1 >= I49 - 1
33.88/33.45	
33.88/33.45	We remove all the strictly oriented dependency pairs.
33.88/33.45	
33.88/33.45	DP problem for innermost termination.
33.88/33.45	P =
33.88/33.45	  f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.45	  f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.45	  f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.46	  f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.46	  f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	  f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.46	R = 
33.88/33.46	  f8(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
33.88/33.46	  f7(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.46	  f6(I7, I8, I9, I10, I11, I12, I13) -> f7(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.46	  f5(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.46	  f5(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.46	  f2(I28, I29, I30, I31, I32, I33, I34) -> f5(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.46	  f4(I37, I38, I39, I40, I41, I42, I43) -> f2(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	  f2(I44, I45, I46, I47, I48, I49, I50) -> f4(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.46	  f2(I54, I55, I56, I57, I58, I59, I60) -> f3(I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
33.88/33.46	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I61, I62, I63, I64, I65, I66, I67) 
33.88/33.46	
33.88/33.46	The dependency graph for this problem is:
33.88/33.46	  1 -> 7
33.88/33.46	  2 -> 1
33.88/33.46	  3 -> 2
33.88/33.46	  4 -> 2
33.88/33.46	  6 -> 7
33.88/33.46	  7 -> 6
33.88/33.46	Where:
33.88/33.46	  1) f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.46	  2) f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.46	  3) f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.46	  4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.46	  6) f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	  7) f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.46	
33.88/33.46	We have the following SCCs.
33.88/33.46	  { 6, 7 }
33.88/33.46	
33.88/33.46	DP problem for innermost termination.
33.88/33.46	P =
33.88/33.46	  f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	  f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.46	R = 
33.88/33.46	  f8(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
33.88/33.46	  f7(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.46	  f6(I7, I8, I9, I10, I11, I12, I13) -> f7(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.46	  f5(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.46	  f5(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.46	  f2(I28, I29, I30, I31, I32, I33, I34) -> f5(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.46	  f4(I37, I38, I39, I40, I41, I42, I43) -> f2(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	  f2(I44, I45, I46, I47, I48, I49, I50) -> f4(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.46	  f2(I54, I55, I56, I57, I58, I59, I60) -> f3(I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
33.88/33.46	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I61, I62, I63, I64, I65, I66, I67) 
33.88/33.46	
33.88/33.46	We use the basic value criterion with the projection function NU:
33.88/33.46	  NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = z7
33.88/33.46	  NU[f4#(z1,z2,z3,z4,z5,z6,z7)] = z7
33.88/33.46	
33.88/33.46	This gives the following inequalities:
33.88/33.46	    ==>  I43 (>! \union =) I43
33.88/33.46	  1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50  ==>  I50 >! -1 + I50
33.88/33.46	
33.88/33.46	We remove all the strictly oriented dependency pairs.
33.88/33.46	
33.88/33.46	DP problem for innermost termination.
33.88/33.46	P =
33.88/33.46	  f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	R = 
33.88/33.46	  f8(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
33.88/33.46	  f7(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
33.88/33.46	  f6(I7, I8, I9, I10, I11, I12, I13) -> f7(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1]
33.88/33.46	  f5(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0]
33.88/33.46	  f5(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, I23, I24, I25, I26, I27) [1 <= I23]
33.88/33.46	  f2(I28, I29, I30, I31, I32, I33, I34) -> f5(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35]
33.88/33.46	  f4(I37, I38, I39, I40, I41, I42, I43) -> f2(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	  f2(I44, I45, I46, I47, I48, I49, I50) -> f4(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50]
33.88/33.46	  f2(I54, I55, I56, I57, I58, I59, I60) -> f3(I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
33.88/33.46	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I61, I62, I63, I64, I65, I66, I67) 
33.88/33.46	
33.88/33.46	The dependency graph for this problem is:
33.88/33.46	  6 -> 
33.88/33.46	Where:
33.88/33.46	  6) f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 
33.88/33.46	
33.88/33.46	We have the following SCCs.
33.88/33.46	  
33.88/36.43	EOF