/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 =
  f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
  f9#(I0, I1, I2, I3, I4) -> f4#(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
  f5#(I5, I6, I7, I8, I9) -> f4#(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
  f5#(I11, I12, I13, I14, I15) -> f2#(I11, I12, I13, I14, -1 + I15) [1 <= I15]
  f8#(I16, I17, I18, I19, I20) -> f3#(I16, I17, I18, I19, I20) 
  f3#(I21, I22, I23, I24, I25) -> f8#(I21, I22, I23, I24, I25) 
  f2#(I31, I32, I33, I34, I35) -> f5#(I31, I32, I33, I34, I35) 
  f4#(I36, I37, I38, I39, I40) -> f1#(I36, I37, I38, I39, I40) 
  f1#(I41, I42, I43, I44, I45) -> f3#(I41, I42, I43, I44, I45) [1 <= I44]
  f1#(I46, I47, I48, I49, I50) -> f2#(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]
R = 
  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
  f9(I0, I1, I2, I3, I4) -> f4(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
  f5(I5, I6, I7, I8, I9) -> f4(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
  f5(I11, I12, I13, I14, I15) -> f2(I11, I12, I13, I14, -1 + I15) [1 <= I15]
  f8(I16, I17, I18, I19, I20) -> f3(I16, I17, I18, I19, I20) 
  f3(I21, I22, I23, I24, I25) -> f8(I21, I22, I23, I24, I25) 
  f6(I26, I27, I28, I29, I30) -> f7(I26, I27, I28, I29, I30) 
  f2(I31, I32, I33, I34, I35) -> f5(I31, I32, I33, I34, I35) 
  f4(I36, I37, I38, I39, I40) -> f1(I36, I37, I38, I39, I40) 
  f1(I41, I42, I43, I44, I45) -> f3(I41, I42, I43, I44, I45) [1 <= I44]
  f1(I46, I47, I48, I49, I50) -> f2(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]

The dependency graph for this problem is:
  0 -> 1
  1 -> 7
  2 -> 7
  3 -> 6
  4 -> 5
  5 -> 4
  6 -> 2, 3
  7 -> 8, 9
  8 -> 5
  9 -> 6
Where:
  0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
  1) f9#(I0, I1, I2, I3, I4) -> f4#(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
  2) f5#(I5, I6, I7, I8, I9) -> f4#(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
  3) f5#(I11, I12, I13, I14, I15) -> f2#(I11, I12, I13, I14, -1 + I15) [1 <= I15]
  4) f8#(I16, I17, I18, I19, I20) -> f3#(I16, I17, I18, I19, I20) 
  5) f3#(I21, I22, I23, I24, I25) -> f8#(I21, I22, I23, I24, I25) 
  6) f2#(I31, I32, I33, I34, I35) -> f5#(I31, I32, I33, I34, I35) 
  7) f4#(I36, I37, I38, I39, I40) -> f1#(I36, I37, I38, I39, I40) 
  8) f1#(I41, I42, I43, I44, I45) -> f3#(I41, I42, I43, I44, I45) [1 <= I44]
  9) f1#(I46, I47, I48, I49, I50) -> f2#(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]

We have the following SCCs.
  { 2, 3, 6, 7, 9 }
  { 4, 5 }

DP problem for innermost termination.
P =
  f8#(I16, I17, I18, I19, I20) -> f3#(I16, I17, I18, I19, I20) 
  f3#(I21, I22, I23, I24, I25) -> f8#(I21, I22, I23, I24, I25) 
R = 
  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
  f9(I0, I1, I2, I3, I4) -> f4(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
  f5(I5, I6, I7, I8, I9) -> f4(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
  f5(I11, I12, I13, I14, I15) -> f2(I11, I12, I13, I14, -1 + I15) [1 <= I15]
  f8(I16, I17, I18, I19, I20) -> f3(I16, I17, I18, I19, I20) 
  f3(I21, I22, I23, I24, I25) -> f8(I21, I22, I23, I24, I25) 
  f6(I26, I27, I28, I29, I30) -> f7(I26, I27, I28, I29, I30) 
  f2(I31, I32, I33, I34, I35) -> f5(I31, I32, I33, I34, I35) 
  f4(I36, I37, I38, I39, I40) -> f1(I36, I37, I38, I39, I40) 
  f1(I41, I42, I43, I44, I45) -> f3(I41, I42, I43, I44, I45) [1 <= I44]
  f1(I46, I47, I48, I49, I50) -> f2(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]