0.78/0.86	YES
0.78/0.86	
0.78/0.86	DP problem for innermost termination.
0.78/0.86	P =
0.78/0.86	  init#(x1, x2, x3, x4, x5) -> f3#(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.78/0.86	  f5#(I0, I1, I2, I3, I4) -> f5#(I5, I1, I6, I7, I8) [-1 <= I5 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I5 + 1 <= I0]
0.78/0.86	  f6#(I9, I10, I11, I12, I13) -> f6#(I9 - 1, I14, I15, I16, I17) [y2 <= y1 - 1 /\ 1 <= I9 - 1 /\ -1 <= y2 - 1 /\ -1 <= y3 - 1 /\ -1 <= y1 - 1 /\ I9 - 1 <= I9 - 1 /\ I14 <= I10 /\ 0 <= I10 - 1 /\ 0 <= I14 - 1]
0.78/0.86	  f2#(I18, I19, I20, I21, I22) -> f6#(I23, I24, I25, I26, I27) [0 <= I28 - 1 /\ I28 <= I19 - 1 /\ -1 <= I23 - 1 /\ -1 <= I19 - 1 /\ I24 + 3 <= I18 /\ 3 <= I18 - 1 /\ 0 <= I24 - 1 /\ I21 + 2 <= I18 /\ 0 = I20]
0.78/0.86	  f3#(I29, I30, I31, I32, I33) -> f6#(1, I34, I35, I36, I37) [0 <= I34 - 1 /\ 0 <= I29 - 1 /\ -1 <= I30 - 1 /\ I34 <= I29]
0.78/0.86	  f4#(I38, I39, I40, I41, I42) -> f5#(I43, I40, I44, I45, I46) [I41 + 4 <= I39 /\ I42 + 2 <= I39 /\ 3 <= I43 - 1 /\ 3 <= I39 - 1 /\ 0 <= I38 - 1 /\ 0 <= I40 - 1 /\ I43 <= I39]
0.78/0.86	  f2#(I47, I48, I49, I50, I51) -> f4#(I52, I53, I54, I55, I50) [0 <= I56 - 1 /\ I56 <= I48 - 1 /\ -1 <= I57 - 1 /\ I52 + 3 <= I47 /\ 3 <= I47 - 1 /\ 0 <= I52 - 1 /\ 3 <= I53 - 1 /\ I50 + 2 <= I47 /\ 0 = I49]
0.78/0.86	  f3#(I58, I59, I60, I61, I62) -> f2#(I63, I59, 0, I64, I65) [3 <= I63 - 1 /\ 0 <= I58 - 1]
0.78/0.86	  f1#(I66, I67, I68, I69, I70) -> f2#(I71, I72, 0, I70, I73) [0 = I69 /\ I70 + 2 <= I67 /\ 3 <= I71 - 1 /\ 3 <= I67 - 1 /\ 0 <= I66 - 1 /\ I71 <= I67]
0.78/0.86	R = 
0.78/0.86	  init(x1, x2, x3, x4, x5) -> f3(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.78/0.86	  f5(I0, I1, I2, I3, I4) -> f5(I5, I1, I6, I7, I8) [-1 <= I5 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I5 + 1 <= I0]
0.78/0.86	  f6(I9, I10, I11, I12, I13) -> f6(I9 - 1, I14, I15, I16, I17) [y2 <= y1 - 1 /\ 1 <= I9 - 1 /\ -1 <= y2 - 1 /\ -1 <= y3 - 1 /\ -1 <= y1 - 1 /\ I9 - 1 <= I9 - 1 /\ I14 <= I10 /\ 0 <= I10 - 1 /\ 0 <= I14 - 1]
0.78/0.86	  f2(I18, I19, I20, I21, I22) -> f6(I23, I24, I25, I26, I27) [0 <= I28 - 1 /\ I28 <= I19 - 1 /\ -1 <= I23 - 1 /\ -1 <= I19 - 1 /\ I24 + 3 <= I18 /\ 3 <= I18 - 1 /\ 0 <= I24 - 1 /\ I21 + 2 <= I18 /\ 0 = I20]
0.78/0.86	  f3(I29, I30, I31, I32, I33) -> f6(1, I34, I35, I36, I37) [0 <= I34 - 1 /\ 0 <= I29 - 1 /\ -1 <= I30 - 1 /\ I34 <= I29]
0.78/0.86	  f4(I38, I39, I40, I41, I42) -> f5(I43, I40, I44, I45, I46) [I41 + 4 <= I39 /\ I42 + 2 <= I39 /\ 3 <= I43 - 1 /\ 3 <= I39 - 1 /\ 0 <= I38 - 1 /\ 0 <= I40 - 1 /\ I43 <= I39]
0.78/0.86	  f2(I47, I48, I49, I50, I51) -> f4(I52, I53, I54, I55, I50) [0 <= I56 - 1 /\ I56 <= I48 - 1 /\ -1 <= I57 - 1 /\ I52 + 3 <= I47 /\ 3 <= I47 - 1 /\ 0 <= I52 - 1 /\ 3 <= I53 - 1 /\ I50 + 2 <= I47 /\ 0 = I49]
0.78/0.86	  f3(I58, I59, I60, I61, I62) -> f2(I63, I59, 0, I64, I65) [3 <= I63 - 1 /\ 0 <= I58 - 1]
0.78/0.86	  f1(I66, I67, I68, I69, I70) -> f2(I71, I72, 0, I70, I73) [0 = I69 /\ I70 + 2 <= I67 /\ 3 <= I71 - 1 /\ 3 <= I67 - 1 /\ 0 <= I66 - 1 /\ I71 <= I67]
0.78/0.86	
0.78/0.86	The dependency graph for this problem is:
0.78/0.86	  0 -> 4, 7
0.78/0.86	  1 -> 1
0.78/0.86	  2 -> 2
0.78/0.86	  3 -> 2
0.78/0.86	  4 -> 
0.78/0.86	  5 -> 1
0.78/0.86	  6 -> 5
0.78/0.86	  7 -> 3, 6
0.78/0.86	  8 -> 3, 6
0.78/0.86	Where:
0.78/0.86	  0) init#(x1, x2, x3, x4, x5) -> f3#(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.78/0.86	  1) f5#(I0, I1, I2, I3, I4) -> f5#(I5, I1, I6, I7, I8) [-1 <= I5 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I5 + 1 <= I0]
0.78/0.86	  2) f6#(I9, I10, I11, I12, I13) -> f6#(I9 - 1, I14, I15, I16, I17) [y2 <= y1 - 1 /\ 1 <= I9 - 1 /\ -1 <= y2 - 1 /\ -1 <= y3 - 1 /\ -1 <= y1 - 1 /\ I9 - 1 <= I9 - 1 /\ I14 <= I10 /\ 0 <= I10 - 1 /\ 0 <= I14 - 1]
0.78/0.86	  3) f2#(I18, I19, I20, I21, I22) -> f6#(I23, I24, I25, I26, I27) [0 <= I28 - 1 /\ I28 <= I19 - 1 /\ -1 <= I23 - 1 /\ -1 <= I19 - 1 /\ I24 + 3 <= I18 /\ 3 <= I18 - 1 /\ 0 <= I24 - 1 /\ I21 + 2 <= I18 /\ 0 = I20]
0.78/0.86	  4) f3#(I29, I30, I31, I32, I33) -> f6#(1, I34, I35, I36, I37) [0 <= I34 - 1 /\ 0 <= I29 - 1 /\ -1 <= I30 - 1 /\ I34 <= I29]
0.78/0.86	  5) f4#(I38, I39, I40, I41, I42) -> f5#(I43, I40, I44, I45, I46) [I41 + 4 <= I39 /\ I42 + 2 <= I39 /\ 3 <= I43 - 1 /\ 3 <= I39 - 1 /\ 0 <= I38 - 1 /\ 0 <= I40 - 1 /\ I43 <= I39]
0.78/0.86	  6) f2#(I47, I48, I49, I50, I51) -> f4#(I52, I53, I54, I55, I50) [0 <= I56 - 1 /\ I56 <= I48 - 1 /\ -1 <= I57 - 1 /\ I52 + 3 <= I47 /\ 3 <= I47 - 1 /\ 0 <= I52 - 1 /\ 3 <= I53 - 1 /\ I50 + 2 <= I47 /\ 0 = I49]
0.78/0.86	  7) f3#(I58, I59, I60, I61, I62) -> f2#(I63, I59, 0, I64, I65) [3 <= I63 - 1 /\ 0 <= I58 - 1]
0.78/0.86	  8) f1#(I66, I67, I68, I69, I70) -> f2#(I71, I72, 0, I70, I73) [0 = I69 /\ I70 + 2 <= I67 /\ 3 <= I71 - 1 /\ 3 <= I67 - 1 /\ 0 <= I66 - 1 /\ I71 <= I67]
0.78/0.86	
0.78/0.86	We have the following SCCs.
0.78/0.86	  { 1 }
0.78/0.86	  { 2 }
0.78/0.86	
0.78/0.86	DP problem for innermost termination.
0.78/0.86	P =
0.78/0.86	  f6#(I9, I10, I11, I12, I13) -> f6#(I9 - 1, I14, I15, I16, I17) [y2 <= y1 - 1 /\ 1 <= I9 - 1 /\ -1 <= y2 - 1 /\ -1 <= y3 - 1 /\ -1 <= y1 - 1 /\ I9 - 1 <= I9 - 1 /\ I14 <= I10 /\ 0 <= I10 - 1 /\ 0 <= I14 - 1]
0.78/0.86	R = 
0.78/0.86	  init(x1, x2, x3, x4, x5) -> f3(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.78/0.86	  f5(I0, I1, I2, I3, I4) -> f5(I5, I1, I6, I7, I8) [-1 <= I5 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I5 + 1 <= I0]
0.78/0.86	  f6(I9, I10, I11, I12, I13) -> f6(I9 - 1, I14, I15, I16, I17) [y2 <= y1 - 1 /\ 1 <= I9 - 1 /\ -1 <= y2 - 1 /\ -1 <= y3 - 1 /\ -1 <= y1 - 1 /\ I9 - 1 <= I9 - 1 /\ I14 <= I10 /\ 0 <= I10 - 1 /\ 0 <= I14 - 1]
0.78/0.86	  f2(I18, I19, I20, I21, I22) -> f6(I23, I24, I25, I26, I27) [0 <= I28 - 1 /\ I28 <= I19 - 1 /\ -1 <= I23 - 1 /\ -1 <= I19 - 1 /\ I24 + 3 <= I18 /\ 3 <= I18 - 1 /\ 0 <= I24 - 1 /\ I21 + 2 <= I18 /\ 0 = I20]
0.78/0.86	  f3(I29, I30, I31, I32, I33) -> f6(1, I34, I35, I36, I37) [0 <= I34 - 1 /\ 0 <= I29 - 1 /\ -1 <= I30 - 1 /\ I34 <= I29]
0.78/0.86	  f4(I38, I39, I40, I41, I42) -> f5(I43, I40, I44, I45, I46) [I41 + 4 <= I39 /\ I42 + 2 <= I39 /\ 3 <= I43 - 1 /\ 3 <= I39 - 1 /\ 0 <= I38 - 1 /\ 0 <= I40 - 1 /\ I43 <= I39]
0.78/0.86	  f2(I47, I48, I49, I50, I51) -> f4(I52, I53, I54, I55, I50) [0 <= I56 - 1 /\ I56 <= I48 - 1 /\ -1 <= I57 - 1 /\ I52 + 3 <= I47 /\ 3 <= I47 - 1 /\ 0 <= I52 - 1 /\ 3 <= I53 - 1 /\ I50 + 2 <= I47 /\ 0 = I49]
0.78/0.86	  f3(I58, I59, I60, I61, I62) -> f2(I63, I59, 0, I64, I65) [3 <= I63 - 1 /\ 0 <= I58 - 1]
0.78/0.86	  f1(I66, I67, I68, I69, I70) -> f2(I71, I72, 0, I70, I73) [0 = I69 /\ I70 + 2 <= I67 /\ 3 <= I71 - 1 /\ 3 <= I67 - 1 /\ 0 <= I66 - 1 /\ I71 <= I67]
0.78/0.86	
0.78/0.86	We use the basic value criterion with the projection function NU:
0.78/0.86	  NU[f6#(z1,z2,z3,z4,z5)] = z1
0.78/0.86	
0.78/0.86	This gives the following inequalities:
0.78/0.86	  y2 <= y1 - 1 /\ 1 <= I9 - 1 /\ -1 <= y2 - 1 /\ -1 <= y3 - 1 /\ -1 <= y1 - 1 /\ I9 - 1 <= I9 - 1 /\ I14 <= I10 /\ 0 <= I10 - 1 /\ 0 <= I14 - 1  ==>  I9 >! I9 - 1
0.78/0.86	
0.78/0.86	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.78/0.86	
0.78/0.86	DP problem for innermost termination.
0.78/0.86	P =
0.78/0.86	  f5#(I0, I1, I2, I3, I4) -> f5#(I5, I1, I6, I7, I8) [-1 <= I5 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I5 + 1 <= I0]
0.78/0.86	R = 
0.78/0.86	  init(x1, x2, x3, x4, x5) -> f3(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.78/0.86	  f5(I0, I1, I2, I3, I4) -> f5(I5, I1, I6, I7, I8) [-1 <= I5 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I5 + 1 <= I0]
0.78/0.86	  f6(I9, I10, I11, I12, I13) -> f6(I9 - 1, I14, I15, I16, I17) [y2 <= y1 - 1 /\ 1 <= I9 - 1 /\ -1 <= y2 - 1 /\ -1 <= y3 - 1 /\ -1 <= y1 - 1 /\ I9 - 1 <= I9 - 1 /\ I14 <= I10 /\ 0 <= I10 - 1 /\ 0 <= I14 - 1]
0.78/0.86	  f2(I18, I19, I20, I21, I22) -> f6(I23, I24, I25, I26, I27) [0 <= I28 - 1 /\ I28 <= I19 - 1 /\ -1 <= I23 - 1 /\ -1 <= I19 - 1 /\ I24 + 3 <= I18 /\ 3 <= I18 - 1 /\ 0 <= I24 - 1 /\ I21 + 2 <= I18 /\ 0 = I20]
0.78/0.86	  f3(I29, I30, I31, I32, I33) -> f6(1, I34, I35, I36, I37) [0 <= I34 - 1 /\ 0 <= I29 - 1 /\ -1 <= I30 - 1 /\ I34 <= I29]
0.78/0.86	  f4(I38, I39, I40, I41, I42) -> f5(I43, I40, I44, I45, I46) [I41 + 4 <= I39 /\ I42 + 2 <= I39 /\ 3 <= I43 - 1 /\ 3 <= I39 - 1 /\ 0 <= I38 - 1 /\ 0 <= I40 - 1 /\ I43 <= I39]
0.78/0.86	  f2(I47, I48, I49, I50, I51) -> f4(I52, I53, I54, I55, I50) [0 <= I56 - 1 /\ I56 <= I48 - 1 /\ -1 <= I57 - 1 /\ I52 + 3 <= I47 /\ 3 <= I47 - 1 /\ 0 <= I52 - 1 /\ 3 <= I53 - 1 /\ I50 + 2 <= I47 /\ 0 = I49]
0.78/0.86	  f3(I58, I59, I60, I61, I62) -> f2(I63, I59, 0, I64, I65) [3 <= I63 - 1 /\ 0 <= I58 - 1]
0.78/0.86	  f1(I66, I67, I68, I69, I70) -> f2(I71, I72, 0, I70, I73) [0 = I69 /\ I70 + 2 <= I67 /\ 3 <= I71 - 1 /\ 3 <= I67 - 1 /\ 0 <= I66 - 1 /\ I71 <= I67]
0.78/0.86	
0.78/0.86	We use the basic value criterion with the projection function NU:
0.78/0.86	  NU[f5#(z1,z2,z3,z4,z5)] = z1
0.78/0.86	
0.78/0.86	This gives the following inequalities:
0.78/0.86	  -1 <= I5 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I5 + 1 <= I0  ==>  I0 >! I5
0.78/0.86	
0.78/0.86	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.78/3.85	EOF