/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 =
  f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 
  f7#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  f5#(I5, I6, I7, I8, I9) -> f3#(I5, 1 + I6, I7, I8, I9) 
  f3#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) 
  f4#(I20, I21, I22, I23, I24) -> f2#(I20, I21, I22, I23, I24) [I22 <= 0]
  f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) [1 <= I27]
  f2#(I30, I31, I32, I33, I34) -> f3#(I30, 0, I32, I33, I34) 
  f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, I37, I38, I39) [I35 <= I37]
  f1#(I40, I41, I42, I43, I44) -> f2#(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40]
R = 
  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
  f7(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  f5(I5, I6, I7, I8, I9) -> f3(I5, 1 + I6, I7, I8, I9) 
  f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, I13, I14) 
  f3(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) 
  f4(I20, I21, I22, I23, I24) -> f2(I20, I21, I22, I23, I24) [I22 <= 0]
  f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) [1 <= I27]
  f2(I30, I31, I32, I33, I34) -> f3(I30, 0, I32, I33, I34) 
  f1(I35, I36, I37, I38, I39) -> f2(I35, I36, I37, I38, I39) [I35 <= I37]
  f1(I40, I41, I42, I43, I44) -> f2(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40]

The dependency graph for this problem is:
  0 -> 1
  1 -> 4, 5
  2 -> 3
  3 -> 2
  4 -> 6
  5 -> 7, 8
  6 -> 3
  7 -> 6
  8 -> 6
Where:
  0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 
  1) f7#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  2) f5#(I5, I6, I7, I8, I9) -> f3#(I5, 1 + I6, I7, I8, I9) 
  3) f3#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) 
  4) f4#(I20, I21, I22, I23, I24) -> f2#(I20, I21, I22, I23, I24) [I22 <= 0]
  5) f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) [1 <= I27]
  6) f2#(I30, I31, I32, I33, I34) -> f3#(I30, 0, I32, I33, I34) 
  7) f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, I37, I38, I39) [I35 <= I37]
  8) f1#(I40, I41, I42, I43, I44) -> f2#(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f5#(I5, I6, I7, I8, I9) -> f3#(I5, 1 + I6, I7, I8, I9) 
  f3#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) 
R = 
  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
  f7(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  f5(I5, I6, I7, I8, I9) -> f3(I5, 1 + I6, I7, I8, I9) 
  f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, I13, I14) 
  f3(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) 
  f4(I20, I21, I22, I23, I24) -> f2(I20, I21, I22, I23, I24) [I22 <= 0]
  f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) [1 <= I27]
  f2(I30, I31, I32, I33, I34) -> f3(I30, 0, I32, I33, I34) 
  f1(I35, I36, I37, I38, I39) -> f2(I35, I36, I37, I38, I39) [I35 <= I37]
  f1(I40, I41, I42, I43, I44) -> f2(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40]