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


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YES

DP problem for innermost termination.
P =
  f13#(x1, x2, x3, x4, x5) -> f12#(x1, x2, x3, x4, x5) 
  f12#(I0, I1, I2, I3, I4) -> f6#(0, I1, I2, I3, I4) 
  f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 
  f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 
  f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 
  f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 
  f8#(I25, I26, I27, I28, I29) -> f7#(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 
  f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 
  f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 
  f3#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  f3#(I55, I56, I57, I58, I59) -> f4#(I55, 100, I57, I58, I59) [100 <= I55]
  f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 
R = 
  f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 
  f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 
  f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 
  f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 
  f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
  f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31]
  f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 
  f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 
  f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 
  f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55]
  f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 8
  2 -> 3
  3 -> 4
  4 -> 8
  5 -> 6
  6 -> 7
  7 -> 9
  8 -> 10, 11
  9 -> 5
  10 -> 12
  11 -> 5
  12 -> 2
Where:
  0) f13#(x1, x2, x3, x4, x5) -> f12#(x1, x2, x3, x4, x5) 
  1) f12#(I0, I1, I2, I3, I4) -> f6#(0, I1, I2, I3, I4) 
  2) f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 
  3) f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 
  4) f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 
  5) f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 
  6) f8#(I25, I26, I27, I28, I29) -> f7#(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  7) f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 
  8) f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 
  9) f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 
  10) f3#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  11) f3#(I55, I56, I57, I58, I59) -> f4#(I55, 100, I57, I58, I59) [100 <= I55]
  12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 

We have the following SCCs.
  { 2, 3, 4, 8, 10, 12 }
  { 5, 6, 7, 9 }

DP problem for innermost termination.
P =
  f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 
  f8#(I25, I26, I27, I28, I29) -> f7#(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 
  f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 
R = 
  f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 
  f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 
  f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 
  f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 
  f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
  f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31]
  f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 
  f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 
  f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 
  f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55]
  f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 

We use the extended value criterion with the projection function NU:
  NU[f5#(x0,x1,x2,x3,x4)] = -x1 + 198
  NU[f7#(x0,x1,x2,x3,x4)] = -x1 + 198
  NU[f8#(x0,x1,x2,x3,x4)] = -x1 + 199
  NU[f4#(x0,x1,x2,x3,x4)] = -x1 + 199

This gives the following inequalities:
    ==>  -I21 + 199 >= -I21 + 199
  1 + I26 <= 200  ==>  -I26 + 199 > -I26 + 198  with  -I26 + 199 >= 0
    ==>  -I36 + 198 >= -I36 + 198
    ==>  -I46 + 198 >= -(1 + I46) + 199

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 
  f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 
  f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 
R = 
  f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 
  f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 
  f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 
  f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 
  f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
  f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31]
  f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 
  f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 
  f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 
  f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55]
  f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 

The dependency graph for this problem is:
  5 -> 
  7 -> 9
  9 -> 5
Where:
  5) f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 
  7) f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 
  9) f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 
  f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 
  f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 
  f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 
  f3#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 
R = 
  f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 
  f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 
  f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 
  f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 
  f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
  f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31]
  f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 
  f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 
  f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 
  f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55]
  f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 

We use the extended value criterion with the projection function NU:
  NU[f1#(x0,x1,x2,x3,x4)] = -x0 + 98
  NU[f3#(x0,x1,x2,x3,x4)] = -x0 + 99
  NU[f6#(x0,x1,x2,x3,x4)] = -x0 + 99
  NU[f10#(x0,x1,x2,x3,x4)] = -x0 + 98
  NU[f11#(x0,x1,x2,x3,x4)] = -x0 + 98
  NU[f2#(x0,x1,x2,x3,x4)] = -x0 + 98

This gives the following inequalities:
    ==>  -I5 + 98 >= -I5 + 98
    ==>  -I10 + 98 >= -I10 + 98
    ==>  -I15 + 98 >= -(1 + I15) + 99
    ==>  -I40 + 99 >= -I40 + 99
  1 + I50 <= 100  ==>  -I50 + 99 > -I50 + 98  with  -I50 + 99 >= 0
    ==>  -I60 + 98 >= -I60 + 98

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 
  f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 
  f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 
  f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 
  f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 
R = 
  f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 
  f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 
  f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 
  f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 
  f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
  f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200]
  f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31]
  f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 
  f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 
  f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 
  f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100]
  f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55]
  f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 

The dependency graph for this problem is:
  2 -> 3
  3 -> 4
  4 -> 8
  8 -> 
  12 -> 2
Where:
  2) f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 
  3) f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 
  4) f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 
  8) f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 
  12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 

We have the following SCCs.