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


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MAYBE

DP problem for innermost termination.
P =
  f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
  f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7, I8) 
  f5#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f6#(I9, I10, I11, I12, rnd5, rnd6, rnd7, I16, I17) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = I11 /\ rnd5 = rnd5]
  f4#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f5#(I18, I19, I20, I21, I22, I23, I24, I25, I26) [I19 = I19]
  f2#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f4#(I27, I28, I29, I30, I31, I29, I30, I34, I35) 
  f1#(I48, I49, I50, I51, I52, I53, I54, I55, I56) -> f2#(I48, I49, I50, I51, I52, I53, I54, rnd8, rnd9) [rnd8 = rnd8 /\ rnd9 = rnd9]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f1(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
  f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f2(I0, I1, I2, I3, I4, I5, I6, I7, I8) 
  f5(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f6(I9, I10, I11, I12, rnd5, rnd6, rnd7, I16, I17) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = I11 /\ rnd5 = rnd5]
  f4(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f5(I18, I19, I20, I21, I22, I23, I24, I25, I26) [I19 = I19]
  f2(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f4(I27, I28, I29, I30, I31, I29, I30, I34, I35) 
  f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f3(rnd1, I37, I38, I39, I40, I45, I46, I43, I44) [I47 = I38 /\ y2 = I39 /\ I47 <= y2 /\ y2 <= I47 /\ I45 = I45 /\ I46 = I46 /\ rnd1 = rnd1]
  f1(I48, I49, I50, I51, I52, I53, I54, I55, I56) -> f2(I48, I49, I50, I51, I52, I53, I54, rnd8, rnd9) [rnd8 = rnd8 /\ rnd9 = rnd9]

The dependency graph for this problem is:
  0 -> 5
  1 -> 4
  2 -> 1
  3 -> 2
  4 -> 3
  5 -> 4
Where:
  0) f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
  1) f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7, I8) 
  2) f5#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f6#(I9, I10, I11, I12, rnd5, rnd6, rnd7, I16, I17) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = I11 /\ rnd5 = rnd5]
  3) f4#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f5#(I18, I19, I20, I21, I22, I23, I24, I25, I26) [I19 = I19]
  4) f2#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f4#(I27, I28, I29, I30, I31, I29, I30, I34, I35) 
  5) f1#(I48, I49, I50, I51, I52, I53, I54, I55, I56) -> f2#(I48, I49, I50, I51, I52, I53, I54, rnd8, rnd9) [rnd8 = rnd8 /\ rnd9 = rnd9]

We have the following SCCs.
  { 1, 2, 3, 4 }

DP problem for innermost termination.
P =
  f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7, I8) 
  f5#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f6#(I9, I10, I11, I12, rnd5, rnd6, rnd7, I16, I17) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = I11 /\ rnd5 = rnd5]
  f4#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f5#(I18, I19, I20, I21, I22, I23, I24, I25, I26) [I19 = I19]
  f2#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f4#(I27, I28, I29, I30, I31, I29, I30, I34, I35) 
R = 
  f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f1(x1, x2, x3, x4, x5, x6, x7, x8, x9) 
  f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f2(I0, I1, I2, I3, I4, I5, I6, I7, I8) 
  f5(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f6(I9, I10, I11, I12, rnd5, rnd6, rnd7, I16, I17) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = I11 /\ rnd5 = rnd5]
  f4(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f5(I18, I19, I20, I21, I22, I23, I24, I25, I26) [I19 = I19]
  f2(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f4(I27, I28, I29, I30, I31, I29, I30, I34, I35) 
  f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f3(rnd1, I37, I38, I39, I40, I45, I46, I43, I44) [I47 = I38 /\ y2 = I39 /\ I47 <= y2 /\ y2 <= I47 /\ I45 = I45 /\ I46 = I46 /\ rnd1 = rnd1]
  f1(I48, I49, I50, I51, I52, I53, I54, I55, I56) -> f2(I48, I49, I50, I51, I52, I53, I54, rnd8, rnd9) [rnd8 = rnd8 /\ rnd9 = rnd9]