4.28/4.69	YES
4.28/4.69	
4.28/4.69	DP problem for innermost termination.
4.28/4.69	P =
4.28/4.69	  f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) 
4.28/4.69	  f10#(I0, I1, I2, I3) -> f4#(1, I1, I2, I3) 
4.28/4.69	  f2#(I4, I5, I6, I7) -> f5#(I4, I5, 999, I7) 
4.28/4.69	  f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6#(I12, I13, I14, I15) -> f9#(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
4.28/4.69	  f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
4.28/4.69	  f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
4.28/4.69	  f4#(I36, I37, I38, I39) -> f3#(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4#(I40, I41, I42, I43) -> f2#(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4#(I44, I45, I46, I47) -> f2#(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	R = 
4.28/4.69	  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
4.28/4.69	  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
4.28/4.69	  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
4.28/4.69	  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
4.28/4.69	  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
4.28/4.69	  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
4.28/4.69	  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
4.28/4.69	  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	The dependency graph for this problem is:
4.28/4.69	  0 -> 1
4.28/4.69	  1 -> 11
4.28/4.69	  2 -> 7
4.28/4.69	  3 -> 7
4.28/4.69	  4 -> 5
4.28/4.69	  5 -> 6
4.28/4.69	  6 -> 5
4.28/4.69	  7 -> 3, 4
4.28/4.69	  8 -> 12, 13
4.28/4.69	  9 -> 8
4.28/4.69	  10 -> 2
4.28/4.69	  11 -> 2
4.28/4.69	  12 -> 8
4.28/4.69	  13 -> 2
4.28/4.69	Where:
4.28/4.69	  0) f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) 
4.28/4.69	  1) f10#(I0, I1, I2, I3) -> f4#(1, I1, I2, I3) 
4.28/4.69	  2) f2#(I4, I5, I6, I7) -> f5#(I4, I5, 999, I7) 
4.28/4.69	  3) f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  4) f6#(I12, I13, I14, I15) -> f9#(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  5) f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
4.28/4.69	  6) f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  7) f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
4.28/4.69	  8) f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
4.28/4.69	  9) f4#(I36, I37, I38, I39) -> f3#(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  10) f4#(I40, I41, I42, I43) -> f2#(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  11) f4#(I44, I45, I46, I47) -> f2#(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  12) f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  13) f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	We have the following SCCs.
4.28/4.69	  { 8, 12 }
4.28/4.69	  { 3, 7 }
4.28/4.69	  { 5, 6 }
4.28/4.69	
4.28/4.69	DP problem for innermost termination.
4.28/4.69	P =
4.28/4.69	  f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
4.28/4.69	  f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	R = 
4.28/4.69	  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
4.28/4.69	  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
4.28/4.69	  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
4.28/4.69	  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
4.28/4.69	  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
4.28/4.69	  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
4.28/4.69	  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
4.28/4.69	  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	We use the basic value criterion with the projection function NU:
4.28/4.69	  NU[f7#(z1,z2,z3,z4)] = z4
4.28/4.69	  NU[f9#(z1,z2,z3,z4)] = z4
4.28/4.69	
4.28/4.69	This gives the following inequalities:
4.28/4.69	    ==>  I19 (>! \union =) I19
4.28/4.69	  0 <= I23  ==>  I23 >! -1 + I23
4.28/4.69	
4.28/4.69	We remove all the strictly oriented dependency pairs.
4.28/4.69	
4.28/4.69	DP problem for innermost termination.
4.28/4.69	P =
4.28/4.69	  f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
4.28/4.69	R = 
4.28/4.69	  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
4.28/4.69	  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
4.28/4.69	  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
4.28/4.69	  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
4.28/4.69	  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
4.28/4.69	  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
4.28/4.69	  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
4.28/4.69	  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	The dependency graph for this problem is:
4.28/4.69	  5 -> 
4.28/4.69	Where:
4.28/4.69	  5) f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
4.28/4.69	
4.28/4.69	We have the following SCCs.
4.28/4.69	  
4.28/4.69	
4.28/4.69	DP problem for innermost termination.
4.28/4.69	P =
4.28/4.69	  f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
4.28/4.69	R = 
4.28/4.69	  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
4.28/4.69	  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
4.28/4.69	  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
4.28/4.69	  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
4.28/4.69	  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
4.28/4.69	  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
4.28/4.69	  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
4.28/4.69	  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	We use the basic value criterion with the projection function NU:
4.28/4.69	  NU[f5#(z1,z2,z3,z4)] = z3
4.28/4.69	  NU[f6#(z1,z2,z3,z4)] = z3
4.28/4.69	
4.28/4.69	This gives the following inequalities:
4.28/4.69	  0 <= I10  ==>  I10 >! -1 + I10
4.28/4.69	    ==>  I30 (>! \union =) I30
4.28/4.69	
4.28/4.69	We remove all the strictly oriented dependency pairs.
4.28/4.69	
4.28/4.69	DP problem for innermost termination.
4.28/4.69	P =
4.28/4.69	  f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
4.28/4.69	R = 
4.28/4.69	  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
4.28/4.69	  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
4.28/4.69	  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
4.28/4.69	  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
4.28/4.69	  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
4.28/4.69	  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
4.28/4.69	  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
4.28/4.69	  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	The dependency graph for this problem is:
4.28/4.69	  7 -> 
4.28/4.69	Where:
4.28/4.69	  7) f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
4.28/4.69	
4.28/4.69	We have the following SCCs.
4.28/4.69	  
4.28/4.69	
4.28/4.69	DP problem for innermost termination.
4.28/4.69	P =
4.28/4.69	  f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
4.28/4.69	  f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	R = 
4.28/4.69	  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
4.28/4.69	  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
4.28/4.69	  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
4.28/4.69	  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
4.28/4.69	  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
4.28/4.69	  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
4.28/4.69	  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
4.28/4.69	  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	We use the basic value criterion with the projection function NU:
4.28/4.69	  NU[f1#(z1,z2,z3,z4)] = z2
4.28/4.69	  NU[f3#(z1,z2,z3,z4)] = z2
4.28/4.69	
4.28/4.69	This gives the following inequalities:
4.28/4.69	    ==>  I33 (>! \union =) I33
4.28/4.69	  0 <= I49  ==>  I49 >! -1 + I49
4.28/4.69	
4.28/4.69	We remove all the strictly oriented dependency pairs.
4.28/4.69	
4.28/4.69	DP problem for innermost termination.
4.28/4.69	P =
4.28/4.69	  f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
4.28/4.69	R = 
4.28/4.69	  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
4.28/4.69	  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
4.28/4.69	  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
4.28/4.69	  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
4.28/4.69	  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
4.28/4.69	  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
4.28/4.69	  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
4.28/4.69	  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
4.28/4.69	  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
4.28/4.69	  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
4.28/4.69	  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
4.28/4.69	  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
4.28/4.69	  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
4.28/4.69	  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
4.28/4.69	  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]
4.28/4.69	
4.28/4.69	The dependency graph for this problem is:
4.28/4.69	  8 -> 
4.28/4.69	Where:
4.28/4.69	  8) f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
4.28/4.69	
4.28/4.69	We have the following SCCs.
4.28/4.69	  
4.28/7.67	EOF