/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) 
  f10#(I0, I1, I2, I3) -> f4#(1, I1, I2, I3) 
  f2#(I4, I5, I6, I7) -> f5#(I4, I5, 999, I7) 
  f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10]
  f6#(I12, I13, I14, I15) -> f9#(I12, I13, I14, 999) [1 + I14 <= 0]
  f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
  f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23]
  f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
  f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
  f4#(I36, I37, I38, I39) -> f3#(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4#(I40, I41, I42, I43) -> f2#(I40, I41, I42, I43) [1 + I40 <= 0]
  f4#(I44, I45, I46, I47) -> f2#(I44, I45, I46, I47) [1 <= I44]
  f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49]
  f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) [1 + I53 <= 0]
R = 
  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 11
  2 -> 7
  3 -> 7
  4 -> 5
  5 -> 6
  6 -> 5
  7 -> 3, 4
  8 -> 12, 13
  9 -> 8
  10 -> 2
  11 -> 2
  12 -> 8
  13 -> 2
Where:
  0) f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) 
  1) f10#(I0, I1, I2, I3) -> f4#(1, I1, I2, I3) 
  2) f2#(I4, I5, I6, I7) -> f5#(I4, I5, 999, I7) 
  3) f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10]
  4) f6#(I12, I13, I14, I15) -> f9#(I12, I13, I14, 999) [1 + I14 <= 0]
  5) f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
  6) f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23]
  7) f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
  8) f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
  9) f4#(I36, I37, I38, I39) -> f3#(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  10) f4#(I40, I41, I42, I43) -> f2#(I40, I41, I42, I43) [1 + I40 <= 0]
  11) f4#(I44, I45, I46, I47) -> f2#(I44, I45, I46, I47) [1 <= I44]
  12) f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49]
  13) f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) [1 + I53 <= 0]

We have the following SCCs.
  { 8, 12 }
  { 3, 7 }
  { 5, 6 }

DP problem for innermost termination.
P =
  f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
  f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23]
R = 
  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]

We use the basic value criterion with the projection function NU:
  NU[f7#(z1,z2,z3,z4)] = z4
  NU[f9#(z1,z2,z3,z4)] = z4

This gives the following inequalities:
    ==>  I19 (>! \union =) I19
  0 <= I23  ==>  I23 >! -1 + I23

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 
R = 
  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]

The dependency graph for this problem is:
  5 -> 
Where:
  5) f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10]
  f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
R = 
  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]

We use the basic value criterion with the projection function NU:
  NU[f5#(z1,z2,z3,z4)] = z3
  NU[f6#(z1,z2,z3,z4)] = z3

This gives the following inequalities:
  0 <= I10  ==>  I10 >! -1 + I10
    ==>  I30 (>! \union =) I30

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 
R = 
  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]

The dependency graph for this problem is:
  7 -> 
Where:
  7) f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
  f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49]
R = 
  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4)] = z2
  NU[f3#(z1,z2,z3,z4)] = z2

This gives the following inequalities:
    ==>  I33 (>! \union =) I33
  0 <= I49  ==>  I49 >! -1 + I49

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 
R = 
  f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 
  f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) 
  f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10]
  f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0]
  f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) 
  f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23]
  f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0]
  f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) 
  f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) 
  f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0]
  f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0]
  f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44]
  f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49]
  f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0]

The dependency graph for this problem is:
  8 -> 
Where:
  8) f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 

We have the following SCCs.