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


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MAYBE

DP problem for innermost termination.
P =
  f12#(x1, x2, x3, x4, x5) -> f11#(x1, x2, x3, x4, x5) 
  f11#(I0, I1, I2, I3, I4) -> f4#(I3, 0, I2, I3, I4) 
  f5#(I5, I6, I7, I8, I9) -> f7#(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f5#(I10, I11, I12, I13, I14) -> f4#(I10, I11, I12, I13, I14) [I11 <= 0]
  f8#(I15, I16, I17, I18, I19) -> f10#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I22 <= 0]
  f10#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10#(I30, I31, I32, I33, I34) -> f9#(I30, I31, I32, I33, I34) [1 <= I34]
  f10#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9#(I40, I41, I42, I43, I44) -> f6#(-1 + I40, 1 + I41, I42, I43, I44) 
  f7#(I45, I46, I47, I48, I49) -> f8#(I45, I46, I47, I48, I49) 
  f6#(I50, I51, I52, I53, I54) -> f7#(I50, I51, -1 + I52, I53, I54) 
  f3#(I55, I56, I57, I58, I59) -> f5#(I55, I56, I57, I58, I59) 
  f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) 
  f1#(I65, I66, I67, I68, I69) -> f3#(-1 + I65, 1 + I66, I67, I68, I69) [1 <= I65]
R = 
  f12(x1, x2, x3, x4, x5) -> f11(x1, x2, x3, x4, x5) 
  f11(I0, I1, I2, I3, I4) -> f4(I3, 0, I2, I3, I4) 
  f5(I5, I6, I7, I8, I9) -> f7(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f5(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I13, I14) [I11 <= 0]
  f8(I15, I16, I17, I18, I19) -> f10(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I22 <= 0]
  f10(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [1 <= I34]
  f10(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9(I40, I41, I42, I43, I44) -> f6(-1 + I40, 1 + I41, I42, I43, I44) 
  f7(I45, I46, I47, I48, I49) -> f8(I45, I46, I47, I48, I49) 
  f6(I50, I51, I52, I53, I54) -> f7(I50, I51, -1 + I52, I53, I54) 
  f3(I55, I56, I57, I58, I59) -> f5(I55, I56, I57, I58, I59) 
  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) 
  f1(I65, I66, I67, I68, I69) -> f3(-1 + I65, 1 + I66, I67, I68, I69) [1 <= I65]
  f1(I70, I71, I72, I73, I74) -> f2(I70, I71, I72, I73, I74) [I70 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 13
  2 -> 10
  3 -> 13
  4 -> 6, 7, 8
  5 -> 12
  6 -> 9
  7 -> 9
  8 -> 11
  9 -> 11
  10 -> 4, 5
  11 -> 10
  12 -> 2, 3
  13 -> 14
  14 -> 12
Where:
  0) f12#(x1, x2, x3, x4, x5) -> f11#(x1, x2, x3, x4, x5) 
  1) f11#(I0, I1, I2, I3, I4) -> f4#(I3, 0, I2, I3, I4) 
  2) f5#(I5, I6, I7, I8, I9) -> f7#(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  3) f5#(I10, I11, I12, I13, I14) -> f4#(I10, I11, I12, I13, I14) [I11 <= 0]
  4) f8#(I15, I16, I17, I18, I19) -> f10#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  5) f8#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I22 <= 0]
  6) f10#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  7) f10#(I30, I31, I32, I33, I34) -> f9#(I30, I31, I32, I33, I34) [1 <= I34]
  8) f10#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  9) f9#(I40, I41, I42, I43, I44) -> f6#(-1 + I40, 1 + I41, I42, I43, I44) 
  10) f7#(I45, I46, I47, I48, I49) -> f8#(I45, I46, I47, I48, I49) 
  11) f6#(I50, I51, I52, I53, I54) -> f7#(I50, I51, -1 + I52, I53, I54) 
  12) f3#(I55, I56, I57, I58, I59) -> f5#(I55, I56, I57, I58, I59) 
  13) f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) 
  14) f1#(I65, I66, I67, I68, I69) -> f3#(-1 + I65, 1 + I66, I67, I68, I69) [1 <= I65]

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 }

DP problem for innermost termination.
P =
  f5#(I5, I6, I7, I8, I9) -> f7#(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f5#(I10, I11, I12, I13, I14) -> f4#(I10, I11, I12, I13, I14) [I11 <= 0]
  f8#(I15, I16, I17, I18, I19) -> f10#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I22 <= 0]
  f10#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10#(I30, I31, I32, I33, I34) -> f9#(I30, I31, I32, I33, I34) [1 <= I34]
  f10#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9#(I40, I41, I42, I43, I44) -> f6#(-1 + I40, 1 + I41, I42, I43, I44) 
  f7#(I45, I46, I47, I48, I49) -> f8#(I45, I46, I47, I48, I49) 
  f6#(I50, I51, I52, I53, I54) -> f7#(I50, I51, -1 + I52, I53, I54) 
  f3#(I55, I56, I57, I58, I59) -> f5#(I55, I56, I57, I58, I59) 
  f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) 
  f1#(I65, I66, I67, I68, I69) -> f3#(-1 + I65, 1 + I66, I67, I68, I69) [1 <= I65]
R = 
  f12(x1, x2, x3, x4, x5) -> f11(x1, x2, x3, x4, x5) 
  f11(I0, I1, I2, I3, I4) -> f4(I3, 0, I2, I3, I4) 
  f5(I5, I6, I7, I8, I9) -> f7(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f5(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I13, I14) [I11 <= 0]
  f8(I15, I16, I17, I18, I19) -> f10(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I22 <= 0]
  f10(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [1 <= I34]
  f10(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9(I40, I41, I42, I43, I44) -> f6(-1 + I40, 1 + I41, I42, I43, I44) 
  f7(I45, I46, I47, I48, I49) -> f8(I45, I46, I47, I48, I49) 
  f6(I50, I51, I52, I53, I54) -> f7(I50, I51, -1 + I52, I53, I54) 
  f3(I55, I56, I57, I58, I59) -> f5(I55, I56, I57, I58, I59) 
  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) 
  f1(I65, I66, I67, I68, I69) -> f3(-1 + I65, 1 + I66, I67, I68, I69) [1 <= I65]
  f1(I70, I71, I72, I73, I74) -> f2(I70, I71, I72, I73, I74) [I70 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5)] = z1
  NU[f6#(z1,z2,z3,z4,z5)] = z1
  NU[f9#(z1,z2,z3,z4,z5)] = z1
  NU[f3#(z1,z2,z3,z4,z5)] = z1
  NU[f10#(z1,z2,z3,z4,z5)] = z1
  NU[f8#(z1,z2,z3,z4,z5)] = z1
  NU[f4#(z1,z2,z3,z4,z5)] = z1
  NU[f7#(z1,z2,z3,z4,z5)] = z1
  NU[f5#(z1,z2,z3,z4,z5)] = z1

This gives the following inequalities:
  1 <= I6  ==>  I5 >= I5
  I11 <= 0  ==>  I10 >= I10
  rnd5 = rnd5 /\ 1 <= I17  ==>  I15 >= I15
  I22 <= 0  ==>  I20 >= I20
  1 + I29 <= 0  ==>  I25 >= I25
  1 <= I34  ==>  I30 >= I30
  0 <= I39 /\ I39 <= 0  ==>  I35 >= I35
    ==>  I40 >= -1 + I40
    ==>  I45 >= I45
    ==>  I50 >= I50
    ==>  I55 >= I55
    ==>  I60 >= I60
  1 <= I65  ==>  I65 > -1 + I65  with  I65 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f5#(I5, I6, I7, I8, I9) -> f7#(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f5#(I10, I11, I12, I13, I14) -> f4#(I10, I11, I12, I13, I14) [I11 <= 0]
  f8#(I15, I16, I17, I18, I19) -> f10#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I22 <= 0]
  f10#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10#(I30, I31, I32, I33, I34) -> f9#(I30, I31, I32, I33, I34) [1 <= I34]
  f10#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9#(I40, I41, I42, I43, I44) -> f6#(-1 + I40, 1 + I41, I42, I43, I44) 
  f7#(I45, I46, I47, I48, I49) -> f8#(I45, I46, I47, I48, I49) 
  f6#(I50, I51, I52, I53, I54) -> f7#(I50, I51, -1 + I52, I53, I54) 
  f3#(I55, I56, I57, I58, I59) -> f5#(I55, I56, I57, I58, I59) 
  f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) 
R = 
  f12(x1, x2, x3, x4, x5) -> f11(x1, x2, x3, x4, x5) 
  f11(I0, I1, I2, I3, I4) -> f4(I3, 0, I2, I3, I4) 
  f5(I5, I6, I7, I8, I9) -> f7(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f5(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I13, I14) [I11 <= 0]
  f8(I15, I16, I17, I18, I19) -> f10(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I22 <= 0]
  f10(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [1 <= I34]
  f10(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9(I40, I41, I42, I43, I44) -> f6(-1 + I40, 1 + I41, I42, I43, I44) 
  f7(I45, I46, I47, I48, I49) -> f8(I45, I46, I47, I48, I49) 
  f6(I50, I51, I52, I53, I54) -> f7(I50, I51, -1 + I52, I53, I54) 
  f3(I55, I56, I57, I58, I59) -> f5(I55, I56, I57, I58, I59) 
  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) 
  f1(I65, I66, I67, I68, I69) -> f3(-1 + I65, 1 + I66, I67, I68, I69) [1 <= I65]
  f1(I70, I71, I72, I73, I74) -> f2(I70, I71, I72, I73, I74) [I70 <= 0]

The dependency graph for this problem is:
  2 -> 10
  3 -> 13
  4 -> 6, 7, 8
  5 -> 12
  6 -> 9
  7 -> 9
  8 -> 11
  9 -> 11
  10 -> 4, 5
  11 -> 10
  12 -> 2, 3
  13 -> 
Where:
  2) f5#(I5, I6, I7, I8, I9) -> f7#(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  3) f5#(I10, I11, I12, I13, I14) -> f4#(I10, I11, I12, I13, I14) [I11 <= 0]
  4) f8#(I15, I16, I17, I18, I19) -> f10#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  5) f8#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I22 <= 0]
  6) f10#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  7) f10#(I30, I31, I32, I33, I34) -> f9#(I30, I31, I32, I33, I34) [1 <= I34]
  8) f10#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  9) f9#(I40, I41, I42, I43, I44) -> f6#(-1 + I40, 1 + I41, I42, I43, I44) 
  10) f7#(I45, I46, I47, I48, I49) -> f8#(I45, I46, I47, I48, I49) 
  11) f6#(I50, I51, I52, I53, I54) -> f7#(I50, I51, -1 + I52, I53, I54) 
  12) f3#(I55, I56, I57, I58, I59) -> f5#(I55, I56, I57, I58, I59) 
  13) f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) 

We have the following SCCs.
  { 2, 4, 5, 6, 7, 8, 9, 10, 11, 12 }

DP problem for innermost termination.
P =
  f5#(I5, I6, I7, I8, I9) -> f7#(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f8#(I15, I16, I17, I18, I19) -> f10#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I22 <= 0]
  f10#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10#(I30, I31, I32, I33, I34) -> f9#(I30, I31, I32, I33, I34) [1 <= I34]
  f10#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9#(I40, I41, I42, I43, I44) -> f6#(-1 + I40, 1 + I41, I42, I43, I44) 
  f7#(I45, I46, I47, I48, I49) -> f8#(I45, I46, I47, I48, I49) 
  f6#(I50, I51, I52, I53, I54) -> f7#(I50, I51, -1 + I52, I53, I54) 
  f3#(I55, I56, I57, I58, I59) -> f5#(I55, I56, I57, I58, I59) 
R = 
  f12(x1, x2, x3, x4, x5) -> f11(x1, x2, x3, x4, x5) 
  f11(I0, I1, I2, I3, I4) -> f4(I3, 0, I2, I3, I4) 
  f5(I5, I6, I7, I8, I9) -> f7(I5, -1 + I6, -1 + I5, I8, I9) [1 <= I6]
  f5(I10, I11, I12, I13, I14) -> f4(I10, I11, I12, I13, I14) [I11 <= 0]
  f8(I15, I16, I17, I18, I19) -> f10(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ 1 <= I17]
  f8(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I22 <= 0]
  f10(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, I29) [1 + I29 <= 0]
  f10(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [1 <= I34]
  f10(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) [0 <= I39 /\ I39 <= 0]
  f9(I40, I41, I42, I43, I44) -> f6(-1 + I40, 1 + I41, I42, I43, I44) 
  f7(I45, I46, I47, I48, I49) -> f8(I45, I46, I47, I48, I49) 
  f6(I50, I51, I52, I53, I54) -> f7(I50, I51, -1 + I52, I53, I54) 
  f3(I55, I56, I57, I58, I59) -> f5(I55, I56, I57, I58, I59) 
  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) 
  f1(I65, I66, I67, I68, I69) -> f3(-1 + I65, 1 + I66, I67, I68, I69) [1 <= I65]
  f1(I70, I71, I72, I73, I74) -> f2(I70, I71, I72, I73, I74) [I70 <= 0]