/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 =
  f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
  f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4]
  f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, rnd5) [rnd5 = rnd5]
  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I12 = I12]
  f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, I19) [I16 = I16]
  f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I20 = I20]
  f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28]
  f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33]
R = 
  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4]
  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, rnd5) [rnd5 = rnd5]
  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I12 = I12]
  f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, I19) [I16 = I16]
  f2(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I20 = I20]
  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28]
  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33]

The dependency graph for this problem is:
  0 -> 1
  1 -> 5
  2 -> 6, 7
  3 -> 2
  4 -> 3
  5 -> 4
  6 -> 5
  7 -> 5
Where:
  0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
  1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4]
  2) f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, rnd5) [rnd5 = rnd5]
  3) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I12 = I12]
  4) f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, I19) [I16 = I16]
  5) f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I20 = I20]
  6) f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28]
  7) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33]

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7 }

DP problem for innermost termination.
P =
  f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, rnd5) [rnd5 = rnd5]
  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I12 = I12]
  f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, I19) [I16 = I16]
  f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I20 = I20]
  f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28]
  f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33]
R = 
  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4]
  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, rnd5) [rnd5 = rnd5]
  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I12 = I12]
  f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, I19) [I16 = I16]
  f2(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I20 = I20]
  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28]
  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33]