2.96/3.01	YES
2.96/3.01	
2.96/3.01	DP problem for innermost termination.
2.96/3.01	P =
2.96/3.01	  init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7#(I0, I1, I2, I3, I4) -> f7#(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6#(I9, I10, I11, I12, I13) -> f7#(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6#(I18, I19, I20, I21, I22) -> f6#(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5#(I23, I24, I25, I26, I27) -> f6#(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5#(I28, I29, I30, I31, I32) -> f5#(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4#(I42, I43, I44, I45, I46) -> f4#(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3#(I49, I50, I51, I52, I53) -> f4#(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3#(I56, I57, I58, I59, I60) -> f3#(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2#(I64, I65, I66, I67, I68) -> f3#(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2#(I72, I73, I74, I75, I76) -> f2#(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1#(I80, I81, I82, I83, I84) -> f2#(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	R = 
2.96/3.01	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7(I0, I1, I2, I3, I4) -> f7(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6(I9, I10, I11, I12, I13) -> f7(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6(I18, I19, I20, I21, I22) -> f6(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5(I23, I24, I25, I26, I27) -> f6(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5(I28, I29, I30, I31, I32) -> f5(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4(I35, I36, I37, I38, I39) -> f5(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4(I42, I43, I44, I45, I46) -> f4(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3(I49, I50, I51, I52, I53) -> f4(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3(I56, I57, I58, I59, I60) -> f3(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2(I64, I65, I66, I67, I68) -> f3(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2(I72, I73, I74, I75, I76) -> f2(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1(I80, I81, I82, I83, I84) -> f2(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	The dependency graph for this problem is:
2.96/3.01	  0 -> 12
2.96/3.01	  1 -> 1
2.96/3.01	  2 -> 1
2.96/3.01	  3 -> 2, 3
2.96/3.01	  4 -> 2, 3
2.96/3.01	  5 -> 4, 5
2.96/3.01	  6 -> 4, 5
2.96/3.01	  7 -> 6, 7
2.96/3.01	  8 -> 6, 7
2.96/3.01	  9 -> 8, 9
2.96/3.01	  10 -> 8, 9
2.96/3.01	  11 -> 10, 11
2.96/3.01	  12 -> 10, 11
2.96/3.01	Where:
2.96/3.01	  0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  1) f7#(I0, I1, I2, I3, I4) -> f7#(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  2) f6#(I9, I10, I11, I12, I13) -> f7#(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  3) f6#(I18, I19, I20, I21, I22) -> f6#(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  4) f5#(I23, I24, I25, I26, I27) -> f6#(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  5) f5#(I28, I29, I30, I31, I32) -> f5#(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  6) f4#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  7) f4#(I42, I43, I44, I45, I46) -> f4#(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  8) f3#(I49, I50, I51, I52, I53) -> f4#(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  9) f3#(I56, I57, I58, I59, I60) -> f3#(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  10) f2#(I64, I65, I66, I67, I68) -> f3#(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  11) f2#(I72, I73, I74, I75, I76) -> f2#(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  12) f1#(I80, I81, I82, I83, I84) -> f2#(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	We have the following SCCs.
2.96/3.01	  { 11 }
2.96/3.01	  { 9 }
2.96/3.01	  { 7 }
2.96/3.01	  { 5 }
2.96/3.01	  { 3 }
2.96/3.01	  { 1 }
2.96/3.01	
2.96/3.01	DP problem for innermost termination.
2.96/3.01	P =
2.96/3.01	  f7#(I0, I1, I2, I3, I4) -> f7#(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	R = 
2.96/3.01	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7(I0, I1, I2, I3, I4) -> f7(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6(I9, I10, I11, I12, I13) -> f7(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6(I18, I19, I20, I21, I22) -> f6(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5(I23, I24, I25, I26, I27) -> f6(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5(I28, I29, I30, I31, I32) -> f5(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4(I35, I36, I37, I38, I39) -> f5(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4(I42, I43, I44, I45, I46) -> f4(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3(I49, I50, I51, I52, I53) -> f4(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3(I56, I57, I58, I59, I60) -> f3(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2(I64, I65, I66, I67, I68) -> f3(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2(I72, I73, I74, I75, I76) -> f2(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1(I80, I81, I82, I83, I84) -> f2(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	We use the basic value criterion with the projection function NU:
2.96/3.01	  NU[f7#(z1,z2,z3,z4,z5)] = z1
2.96/3.01	
2.96/3.01	This gives the following inequalities:
2.96/3.01	  -1 <= I0 - 1  ==>  I0 >! I0 - 1
2.96/3.01	
2.96/3.01	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.96/3.01	
2.96/3.01	DP problem for innermost termination.
2.96/3.01	P =
2.96/3.01	  f6#(I18, I19, I20, I21, I22) -> f6#(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	R = 
2.96/3.01	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7(I0, I1, I2, I3, I4) -> f7(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6(I9, I10, I11, I12, I13) -> f7(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6(I18, I19, I20, I21, I22) -> f6(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5(I23, I24, I25, I26, I27) -> f6(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5(I28, I29, I30, I31, I32) -> f5(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4(I35, I36, I37, I38, I39) -> f5(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4(I42, I43, I44, I45, I46) -> f4(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3(I49, I50, I51, I52, I53) -> f4(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3(I56, I57, I58, I59, I60) -> f3(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2(I64, I65, I66, I67, I68) -> f3(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2(I72, I73, I74, I75, I76) -> f2(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1(I80, I81, I82, I83, I84) -> f2(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	We use the reverse value criterion with the projection function NU:
2.96/3.01	  NU[f6#(z1,z2,z3,z4,z5)] = z5 + -1 * z4
2.96/3.01	
2.96/3.01	This gives the following inequalities:
2.96/3.01	  -1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19  ==>  I22 + -1 * I21 > I22 + -1 * (I21 + 1)  with  I22 + -1 * I21 >= 0
2.96/3.01	
2.96/3.01	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.96/3.01	
2.96/3.01	DP problem for innermost termination.
2.96/3.01	P =
2.96/3.01	  f5#(I28, I29, I30, I31, I32) -> f5#(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	R = 
2.96/3.01	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7(I0, I1, I2, I3, I4) -> f7(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6(I9, I10, I11, I12, I13) -> f7(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6(I18, I19, I20, I21, I22) -> f6(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5(I23, I24, I25, I26, I27) -> f6(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5(I28, I29, I30, I31, I32) -> f5(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4(I35, I36, I37, I38, I39) -> f5(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4(I42, I43, I44, I45, I46) -> f4(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3(I49, I50, I51, I52, I53) -> f4(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3(I56, I57, I58, I59, I60) -> f3(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2(I64, I65, I66, I67, I68) -> f3(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2(I72, I73, I74, I75, I76) -> f2(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1(I80, I81, I82, I83, I84) -> f2(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	We use the basic value criterion with the projection function NU:
2.96/3.01	  NU[f5#(z1,z2,z3,z4,z5)] = z3
2.96/3.01	
2.96/3.01	This gives the following inequalities:
2.96/3.01	  0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29  ==>  I30 >! I30 - 1
2.96/3.01	
2.96/3.01	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.96/3.01	
2.96/3.01	DP problem for innermost termination.
2.96/3.01	P =
2.96/3.01	  f4#(I42, I43, I44, I45, I46) -> f4#(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	R = 
2.96/3.01	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7(I0, I1, I2, I3, I4) -> f7(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6(I9, I10, I11, I12, I13) -> f7(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6(I18, I19, I20, I21, I22) -> f6(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5(I23, I24, I25, I26, I27) -> f6(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5(I28, I29, I30, I31, I32) -> f5(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4(I35, I36, I37, I38, I39) -> f5(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4(I42, I43, I44, I45, I46) -> f4(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3(I49, I50, I51, I52, I53) -> f4(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3(I56, I57, I58, I59, I60) -> f3(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2(I64, I65, I66, I67, I68) -> f3(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2(I72, I73, I74, I75, I76) -> f2(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1(I80, I81, I82, I83, I84) -> f2(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	We use the reverse value criterion with the projection function NU:
2.96/3.01	  NU[f4#(z1,z2,z3,z4,z5)] = z3 + -1 * z2
2.96/3.01	
2.96/3.01	This gives the following inequalities:
2.96/3.01	  -1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1  ==>  I44 + -1 * I43 > I44 + -1 * (I43 + 1)  with  I44 + -1 * I43 >= 0
2.96/3.01	
2.96/3.01	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.96/3.01	
2.96/3.01	DP problem for innermost termination.
2.96/3.01	P =
2.96/3.01	  f3#(I56, I57, I58, I59, I60) -> f3#(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	R = 
2.96/3.01	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7(I0, I1, I2, I3, I4) -> f7(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6(I9, I10, I11, I12, I13) -> f7(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6(I18, I19, I20, I21, I22) -> f6(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5(I23, I24, I25, I26, I27) -> f6(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5(I28, I29, I30, I31, I32) -> f5(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4(I35, I36, I37, I38, I39) -> f5(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4(I42, I43, I44, I45, I46) -> f4(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3(I49, I50, I51, I52, I53) -> f4(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3(I56, I57, I58, I59, I60) -> f3(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2(I64, I65, I66, I67, I68) -> f3(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2(I72, I73, I74, I75, I76) -> f2(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1(I80, I81, I82, I83, I84) -> f2(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	We use the reverse value criterion with the projection function NU:
2.96/3.01	  NU[f3#(z1,z2,z3,z4,z5)] = z2 + -1 * z1
2.96/3.01	
2.96/3.01	This gives the following inequalities:
2.96/3.01	  0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57  ==>  I57 + -1 * I56 > I57 + -1 * (I56 + 1)  with  I57 + -1 * I56 >= 0
2.96/3.01	
2.96/3.01	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.96/3.01	
2.96/3.01	DP problem for innermost termination.
2.96/3.01	P =
2.96/3.01	  f2#(I72, I73, I74, I75, I76) -> f2#(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	R = 
2.96/3.01	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
2.96/3.01	  f7(I0, I1, I2, I3, I4) -> f7(I0 - 1, I5, I6, I7, I8) [-1 <= I0 - 1]
2.96/3.01	  f6(I9, I10, I11, I12, I13) -> f7(1000 * I9 + 100 * I10 + 10 * I11 + I12, I14, I15, I16, I17) [0 <= 1000 * I9 + 100 * I10 + 10 * I11 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ 0 <= 1000 * I9 + 100 * I10 /\ 0 <= 10 * I11 /\ 0 <= 1000 * I9 /\ I12 <= I13 /\ 0 <= 100 * I10]
2.96/3.01	  f6(I18, I19, I20, I21, I22) -> f6(I18, I19, I20, I21 + 1, I22) [-1 <= I21 - 1 /\ 0 <= 1000 * I18 + 100 * I19 + 10 * I20 /\ -1 <= I20 - 1 /\ 0 <= 1000 * I18 + 100 * I19 /\ 0 <= 10 * I20 /\ 0 <= 1000 * I18 /\ I21 <= I22 /\ 0 <= 100 * I19]
2.96/3.01	  f5(I23, I24, I25, I26, I27) -> f6(I23, I24, I25, 0, 2 * I23 + 3 * I24 + 4 * I25) [0 <= 4 * I25 /\ 0 <= 2 * I23 + 3 * I24 /\ 0 <= 2 * I23 /\ -1 <= I25 - 1 /\ 0 <= 3 * I24]
2.96/3.01	  f5(I28, I29, I30, I31, I32) -> f5(I28, I29, I30 - 1, I33, I34) [0 <= 4 * I30 /\ 0 <= 2 * I28 + 3 * I29 /\ 0 <= 2 * I28 /\ -1 <= I30 - 1 /\ 0 <= 3 * I29]
2.96/3.01	  f4(I35, I36, I37, I38, I39) -> f5(I35, I36, I35 + I36, I40, I41) [-1 <= I36 - 1 /\ I36 <= I37 /\ -1 <= I35 - 1]
2.96/3.01	  f4(I42, I43, I44, I45, I46) -> f4(I42, I43 + 1, I44, I47, I48) [-1 <= I43 - 1 /\ I43 <= I44 /\ -1 <= I42 - 1]
2.96/3.01	  f3(I49, I50, I51, I52, I53) -> f4(I49, 0, 2 * I49, I54, I55) [I49 <= I50 /\ 0 <= I50 - 1]
2.96/3.01	  f3(I56, I57, I58, I59, I60) -> f3(I56 + 1, I57, I61, I62, I63) [0 <= I57 - 1 /\ -1 <= I56 - 1 /\ I56 <= I57]
2.96/3.01	  f2(I64, I65, I66, I67, I68) -> f3(0, I66, I69, I70, I71) [0 <= I64 - 1 /\ 0 <= I66 - 1 /\ I65 <= I66 - 1]
2.96/3.01	  f2(I72, I73, I74, I75, I76) -> f2(I77, I73 + 1, I74, I78, I79) [0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1]
2.96/3.01	  f1(I80, I81, I82, I83, I84) -> f2(I85, 0, I81, I86, I87) [0 <= I85 - 1 /\ 0 <= I80 - 1 /\ -1 <= I81 - 1 /\ I85 <= I80]
2.96/3.01	
2.96/3.01	We use the reverse value criterion with the projection function NU:
2.96/3.01	  NU[f2#(z1,z2,z3,z4,z5)] = z3 - 1 + -1 * z2
2.96/3.01	
2.96/3.01	This gives the following inequalities:
2.96/3.01	  0 <= I77 - 1 /\ 0 <= I72 - 1 /\ I77 <= I72 /\ 0 <= I74 - 1 /\ I73 <= I74 - 1  ==>  I74 - 1 + -1 * I73 > I74 - 1 + -1 * (I73 + 1)  with  I74 - 1 + -1 * I73 >= 0
2.96/3.01	
2.96/3.01	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.96/5.99	EOF