4.80/4.78	YES
4.80/4.78	
4.80/4.78	DP problem for innermost termination.
4.80/4.78	P =
4.80/4.78	  f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
4.80/4.78	  f10#(I0, I1, I2, I3, I4) -> f6#(I0, I1, I2, I3, rnd5) [rnd5 = rnd5]
4.80/4.78	  f2#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [0 <= I5]
4.80/4.78	  f2#(I10, I11, I12, I13, I14) -> f7#(I10, I11, I12, I13, I14) [1 + I10 <= 0]
4.80/4.78	  f9#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, I18, I19) [1 + I15 <= 1023]
4.80/4.78	  f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) [1023 <= I20]
4.80/4.78	  f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) 
4.80/4.78	  f6#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) [1 + I39 <= 0]
4.80/4.78	  f6#(I40, I41, I42, I43, I44) -> f5#(I40, I41, I42, I43, I44) [1 <= I44]
4.80/4.78	  f6#(I45, I46, I47, I48, I49) -> f4#(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0]
4.80/4.78	  f5#(I50, I51, I52, I53, I54) -> f4#(I50, I51, I52, 0, I54) 
4.80/4.78	  f4#(I55, I56, I57, I58, I59) -> f3#(I55, 0, I57, I58, I59) 
4.80/4.78	  f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63]
4.80/4.78	  f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I68 <= I66]
4.80/4.78	R = 
4.80/4.78	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
4.80/4.78	  f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5]
4.80/4.78	  f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5]
4.80/4.78	  f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0]
4.80/4.78	  f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023]
4.80/4.78	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20]
4.80/4.78	  f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) 
4.80/4.78	  f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) 
4.80/4.78	  f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0]
4.80/4.78	  f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44]
4.80/4.78	  f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0]
4.80/4.78	  f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) 
4.80/4.78	  f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) 
4.80/4.78	  f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63]
4.80/4.78	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66]
4.80/4.78	
4.80/4.78	The dependency graph for this problem is:
4.80/4.78	  0 -> 1
4.80/4.78	  1 -> 7, 8, 9
4.80/4.78	  2 -> 4, 5
4.80/4.78	  3 -> 
4.80/4.78	  4 -> 
4.80/4.78	  5 -> 
4.80/4.78	  6 -> 12, 13
4.80/4.78	  7 -> 10
4.80/4.78	  8 -> 10
4.80/4.78	  9 -> 11
4.80/4.78	  10 -> 11
4.80/4.78	  11 -> 6
4.80/4.78	  12 -> 6
4.80/4.78	  13 -> 2, 3
4.80/4.78	Where:
4.80/4.78	  0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
4.80/4.78	  1) f10#(I0, I1, I2, I3, I4) -> f6#(I0, I1, I2, I3, rnd5) [rnd5 = rnd5]
4.80/4.78	  2) f2#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [0 <= I5]
4.80/4.78	  3) f2#(I10, I11, I12, I13, I14) -> f7#(I10, I11, I12, I13, I14) [1 + I10 <= 0]
4.80/4.78	  4) f9#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, I18, I19) [1 + I15 <= 1023]
4.80/4.78	  5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) [1023 <= I20]
4.80/4.78	  6) f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) 
4.80/4.78	  7) f6#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) [1 + I39 <= 0]
4.80/4.78	  8) f6#(I40, I41, I42, I43, I44) -> f5#(I40, I41, I42, I43, I44) [1 <= I44]
4.80/4.78	  9) f6#(I45, I46, I47, I48, I49) -> f4#(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0]
4.80/4.78	  10) f5#(I50, I51, I52, I53, I54) -> f4#(I50, I51, I52, 0, I54) 
4.80/4.78	  11) f4#(I55, I56, I57, I58, I59) -> f3#(I55, 0, I57, I58, I59) 
4.80/4.78	  12) f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63]
4.80/4.78	  13) f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I68 <= I66]
4.80/4.78	
4.80/4.78	We have the following SCCs.
4.80/4.78	  { 6, 12 }
4.80/4.78	
4.80/4.78	DP problem for innermost termination.
4.80/4.78	P =
4.80/4.78	  f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) 
4.80/4.78	  f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63]
4.80/4.78	R = 
4.80/4.78	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
4.80/4.78	  f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5]
4.80/4.78	  f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5]
4.80/4.78	  f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0]
4.80/4.78	  f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023]
4.80/4.78	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20]
4.80/4.78	  f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) 
4.80/4.78	  f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) 
4.80/4.78	  f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0]
4.80/4.78	  f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44]
4.80/4.78	  f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0]
4.80/4.78	  f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) 
4.80/4.78	  f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) 
4.80/4.78	  f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63]
4.80/4.78	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66]
4.80/4.78	
4.80/4.78	We use the reverse value criterion with the projection function NU:
4.80/4.78	  NU[f1#(z1,z2,z3,z4,z5)] = z4 + -1 * z2
4.80/4.78	  NU[f3#(z1,z2,z3,z4,z5)] = z4 + -1 * z2
4.80/4.78	
4.80/4.78	This gives the following inequalities:
4.80/4.78	    ==>  I33 + -1 * I31 >= I33 + -1 * I31
4.80/4.78	  I61 <= I63  ==>  I63 + -1 * I61 > I63 + -1 * (1 + I61)  with  I63 + -1 * I61 >= 0
4.80/4.78	
4.80/4.78	We remove all the strictly oriented dependency pairs.
4.80/4.78	
4.80/4.78	DP problem for innermost termination.
4.80/4.78	P =
4.80/4.78	  f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) 
4.80/4.78	R = 
4.80/4.78	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
4.80/4.78	  f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5]
4.80/4.78	  f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5]
4.80/4.78	  f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0]
4.80/4.78	  f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023]
4.80/4.78	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20]
4.80/4.78	  f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) 
4.80/4.78	  f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) 
4.80/4.78	  f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0]
4.80/4.78	  f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44]
4.80/4.78	  f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0]
4.80/4.78	  f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) 
4.80/4.78	  f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) 
4.80/4.78	  f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63]
4.80/4.78	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66]
4.80/4.78	
4.80/4.78	The dependency graph for this problem is:
4.80/4.78	  6 -> 
4.80/4.78	Where:
4.80/4.78	  6) f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) 
4.80/4.78	
4.80/4.78	We have the following SCCs.
4.80/4.78	  
4.80/7.76	EOF