7.50/7.45	YES
7.50/7.45	
7.50/7.45	DP problem for innermost termination.
7.50/7.45	P =
7.50/7.45	  f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 
7.50/7.45	  f6#(I0, I1, I2, I3, I4, I5) -> f5#(I0, I1, I2, I3, I4, I5) 
7.50/7.45	  f6#(I6, I7, I8, I9, I10, I11) -> f4#(I6, I7, I8, I9, I10, I11) 
7.50/7.45	  f6#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 
7.50/7.45	  f6#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
7.50/7.45	  f6#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
7.50/7.45	  f5#(I36, I37, I38, I39, I40, I41) -> f4#(I40, I41, I42, I43, I44, I45) [I45 = I43 /\ I44 = I42 /\ I43 = I43 /\ I42 = I42]
7.50/7.45	  f4#(I46, I47, I48, I49, I50, I51) -> f3#(I50, I51, I48, I49, I50, I51) [3 <= I50 /\ 1 + I50 <= I51]
7.50/7.45	  f4#(I52, I53, I54, I55, I56, I57) -> f1#(I56, I57, I54, I55, I56, I57) [I57 <= I56]
7.50/7.45	  f4#(I58, I59, I60, I61, I62, I63) -> f1#(I62, I63, I60, I61, I62, I63) [I62 <= 2]
7.50/7.45	  f3#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	R = 
7.50/7.45	  f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 
7.50/7.45	  f6(I0, I1, I2, I3, I4, I5) -> f5(I0, I1, I2, I3, I4, I5) 
7.50/7.45	  f6(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I8, I9, I10, I11) 
7.50/7.45	  f6(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) 
7.50/7.45	  f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
7.50/7.45	  f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) 
7.50/7.45	  f6(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
7.50/7.45	  f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I42, I43, I44, I45) [I45 = I43 /\ I44 = I42 /\ I43 = I43 /\ I42 = I42]
7.50/7.45	  f4(I46, I47, I48, I49, I50, I51) -> f3(I50, I51, I48, I49, I50, I51) [3 <= I50 /\ 1 + I50 <= I51]
7.50/7.45	  f4(I52, I53, I54, I55, I56, I57) -> f1(I56, I57, I54, I55, I56, I57) [I57 <= I56]
7.50/7.45	  f4(I58, I59, I60, I61, I62, I63) -> f1(I62, I63, I60, I61, I62, I63) [I62 <= 2]
7.50/7.45	  f3(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	  f1(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ I77 = I77 /\ I76 = I76]
7.50/7.45	
7.50/7.45	The dependency graph for this problem is:
7.50/7.45	  0 -> 1, 2, 3, 4, 5
7.50/7.45	  1 -> 6
7.50/7.45	  2 -> 7, 8, 9
7.50/7.45	  3 -> 10
7.50/7.45	  4 -> 
7.50/7.45	  5 -> 6
7.50/7.45	  6 -> 7, 8, 9
7.50/7.45	  7 -> 10
7.50/7.45	  8 -> 
7.50/7.45	  9 -> 
7.50/7.45	  10 -> 7, 8, 9
7.50/7.45	Where:
7.50/7.45	  0) f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 
7.50/7.45	  1) f6#(I0, I1, I2, I3, I4, I5) -> f5#(I0, I1, I2, I3, I4, I5) 
7.50/7.45	  2) f6#(I6, I7, I8, I9, I10, I11) -> f4#(I6, I7, I8, I9, I10, I11) 
7.50/7.45	  3) f6#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 
7.50/7.45	  4) f6#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
7.50/7.45	  5) f6#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
7.50/7.45	  6) f5#(I36, I37, I38, I39, I40, I41) -> f4#(I40, I41, I42, I43, I44, I45) [I45 = I43 /\ I44 = I42 /\ I43 = I43 /\ I42 = I42]
7.50/7.45	  7) f4#(I46, I47, I48, I49, I50, I51) -> f3#(I50, I51, I48, I49, I50, I51) [3 <= I50 /\ 1 + I50 <= I51]
7.50/7.45	  8) f4#(I52, I53, I54, I55, I56, I57) -> f1#(I56, I57, I54, I55, I56, I57) [I57 <= I56]
7.50/7.45	  9) f4#(I58, I59, I60, I61, I62, I63) -> f1#(I62, I63, I60, I61, I62, I63) [I62 <= 2]
7.50/7.45	  10) f3#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	
7.50/7.45	We have the following SCCs.
7.50/7.45	  { 7, 10 }
7.50/7.45	
7.50/7.45	DP problem for innermost termination.
7.50/7.45	P =
7.50/7.45	  f4#(I46, I47, I48, I49, I50, I51) -> f3#(I50, I51, I48, I49, I50, I51) [3 <= I50 /\ 1 + I50 <= I51]
7.50/7.45	  f3#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	R = 
7.50/7.45	  f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 
7.50/7.45	  f6(I0, I1, I2, I3, I4, I5) -> f5(I0, I1, I2, I3, I4, I5) 
7.50/7.45	  f6(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I8, I9, I10, I11) 
7.50/7.45	  f6(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) 
7.50/7.45	  f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
7.50/7.45	  f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) 
7.50/7.45	  f6(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
7.50/7.45	  f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I42, I43, I44, I45) [I45 = I43 /\ I44 = I42 /\ I43 = I43 /\ I42 = I42]
7.50/7.45	  f4(I46, I47, I48, I49, I50, I51) -> f3(I50, I51, I48, I49, I50, I51) [3 <= I50 /\ 1 + I50 <= I51]
7.50/7.45	  f4(I52, I53, I54, I55, I56, I57) -> f1(I56, I57, I54, I55, I56, I57) [I57 <= I56]
7.50/7.45	  f4(I58, I59, I60, I61, I62, I63) -> f1(I62, I63, I60, I61, I62, I63) [I62 <= 2]
7.50/7.45	  f3(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	  f1(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ I77 = I77 /\ I76 = I76]
7.50/7.45	
7.50/7.45	We use the reverse value criterion with the projection function NU:
7.50/7.45	  NU[f3#(z1,z2,z3,z4,z5,z6)] = 1 + z6 + -1 * (1 + 2 * z5)
7.50/7.45	  NU[f4#(z1,z2,z3,z4,z5,z6)] = z6 + -1 * (1 + z5)
7.50/7.45	
7.50/7.45	This gives the following inequalities:
7.50/7.45	  3 <= I50 /\ 1 + I50 <= I51  ==>  I51 + -1 * (1 + I50) > 1 + I51 + -1 * (1 + 2 * I50)  with  I51 + -1 * (1 + I50) >= 0
7.50/7.45	    ==>  1 + I69 + -1 * (1 + 2 * I68) >= 1 + I69 + -1 * (1 + 2 * I68)
7.50/7.45	
7.50/7.45	We remove all the strictly oriented dependency pairs.
7.50/7.45	
7.50/7.45	DP problem for innermost termination.
7.50/7.45	P =
7.50/7.45	  f3#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	R = 
7.50/7.45	  f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 
7.50/7.45	  f6(I0, I1, I2, I3, I4, I5) -> f5(I0, I1, I2, I3, I4, I5) 
7.50/7.45	  f6(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I8, I9, I10, I11) 
7.50/7.45	  f6(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) 
7.50/7.45	  f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
7.50/7.45	  f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) 
7.50/7.45	  f6(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
7.50/7.45	  f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I42, I43, I44, I45) [I45 = I43 /\ I44 = I42 /\ I43 = I43 /\ I42 = I42]
7.50/7.45	  f4(I46, I47, I48, I49, I50, I51) -> f3(I50, I51, I48, I49, I50, I51) [3 <= I50 /\ 1 + I50 <= I51]
7.50/7.45	  f4(I52, I53, I54, I55, I56, I57) -> f1(I56, I57, I54, I55, I56, I57) [I57 <= I56]
7.50/7.45	  f4(I58, I59, I60, I61, I62, I63) -> f1(I62, I63, I60, I61, I62, I63) [I62 <= 2]
7.50/7.45	  f3(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	  f1(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ I77 = I77 /\ I76 = I76]
7.50/7.45	
7.50/7.45	The dependency graph for this problem is:
7.50/7.45	  10 -> 
7.50/7.45	Where:
7.50/7.45	  10) f3#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, 2 * I68, 1 + I69) 
7.50/7.45	
7.50/7.45	We have the following SCCs.
7.50/7.45	  
7.50/10.42	EOF