/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 =
  f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1#(I0, I1, I2, I3, I4, rnd6, rnd7, rnd8, rnd9, rnd10) [rnd8 = rnd6 /\ rnd7 = rnd10 /\ rnd10 = rnd10 /\ rnd9 = rnd9 /\ rnd6 = rnd6]
  f2#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) 
  f1#(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2#(I30, I31, rnd3, I40, I41, I35, I36, I37, I38, I39) [I42 = I38 /\ y3 = I31 /\ 0 <= -1 + y3 /\ I40 = I40 /\ I41 = I41 /\ I43 = I31 /\ rnd3 = rnd3]
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1(I0, I1, I2, I3, I4, rnd6, rnd7, rnd8, rnd9, rnd10) [rnd8 = rnd6 /\ rnd7 = rnd10 /\ rnd10 = rnd10 /\ rnd9 = rnd9 /\ rnd6 = rnd6]
  f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f3(rnd1, I11, I12, rnd4, rnd5, I15, I16, I17, I18, I19) [y1 = I18 /\ y2 = I11 /\ y2 <= 0 /\ rnd4 = rnd4 /\ rnd5 = rnd5 /\ rnd1 = rnd1]
  f2(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) 
  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, rnd3, I40, I41, I35, I36, I37, I38, I39) [I42 = I38 /\ y3 = I31 /\ 0 <= -1 + y3 /\ I40 = I40 /\ I41 = I41 /\ I43 = I31 /\ rnd3 = rnd3]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1#(I0, I1, I2, I3, I4, rnd6, rnd7, rnd8, rnd9, rnd10) [rnd8 = rnd6 /\ rnd7 = rnd10 /\ rnd10 = rnd10 /\ rnd9 = rnd9 /\ rnd6 = rnd6]
  2) f2#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) 
  3) f1#(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2#(I30, I31, rnd3, I40, I41, I35, I36, I37, I38, I39) [I42 = I38 /\ y3 = I31 /\ 0 <= -1 + y3 /\ I40 = I40 /\ I41 = I41 /\ I43 = I31 /\ rnd3 = rnd3]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f2#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) 
  f1#(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2#(I30, I31, rnd3, I40, I41, I35, I36, I37, I38, I39) [I42 = I38 /\ y3 = I31 /\ 0 <= -1 + y3 /\ I40 = I40 /\ I41 = I41 /\ I43 = I31 /\ rnd3 = rnd3]
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1(I0, I1, I2, I3, I4, rnd6, rnd7, rnd8, rnd9, rnd10) [rnd8 = rnd6 /\ rnd7 = rnd10 /\ rnd10 = rnd10 /\ rnd9 = rnd9 /\ rnd6 = rnd6]
  f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f3(rnd1, I11, I12, rnd4, rnd5, I15, I16, I17, I18, I19) [y1 = I18 /\ y2 = I11 /\ y2 <= 0 /\ rnd4 = rnd4 /\ rnd5 = rnd5 /\ rnd1 = rnd1]
  f2(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) 
  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, rnd3, I40, I41, I35, I36, I37, I38, I39) [I42 = I38 /\ y3 = I31 /\ 0 <= -1 + y3 /\ I40 = I40 /\ I41 = I41 /\ I43 = I31 /\ rnd3 = rnd3]