/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) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8]
  f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
  f4#(I15, I16, I17, I18, I19) -> f1#(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
  f4#(I23, I24, I25, I26, I27) -> f3#(I23, I24, I25, I26, I27) [1 <= I27]
  f8#(I28, I29, I30, I31, I32) -> f7#(I28, I29, I30, I31, I32) 
  f7#(I33, I34, I35, I36, I37) -> f8#(I33, I34, I35, I36, I37) 
  f3#(I43, I44, I45, I46, I47) -> f4#(I43, I44, I45, I46, I47) 
  f1#(I48, I49, I50, I51, I52) -> f2#(I48, I49, I50, I51, I52) 
R = 
  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
  f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8]
  f2(I10, I11, I12, I13, I14) -> f3(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
  f4(I15, I16, I17, I18, I19) -> f1(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
  f4(I23, I24, I25, I26, I27) -> f3(I23, I24, I25, I26, I27) [1 <= I27]
  f8(I28, I29, I30, I31, I32) -> f7(I28, I29, I30, I31, I32) 
  f7(I33, I34, I35, I36, I37) -> f8(I33, I34, I35, I36, I37) 
  f5(I38, I39, I40, I41, I42) -> f6(I38, I39, I40, I41, I42) 
  f3(I43, I44, I45, I46, I47) -> f4(I43, I44, I45, I46, I47) 
  f1(I48, I49, I50, I51, I52) -> f2(I48, I49, I50, I51, I52) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 9
  2 -> 7
  3 -> 8
  4 -> 9
  5 -> 8
  6 -> 7
  7 -> 6
  8 -> 4, 5
  9 -> 2, 3
Where:
  0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
  1) f9#(I0, I1, I2, I3, I4) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  2) f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8]
  3) f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
  4) f4#(I15, I16, I17, I18, I19) -> f1#(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
  5) f4#(I23, I24, I25, I26, I27) -> f3#(I23, I24, I25, I26, I27) [1 <= I27]
  6) f8#(I28, I29, I30, I31, I32) -> f7#(I28, I29, I30, I31, I32) 
  7) f7#(I33, I34, I35, I36, I37) -> f8#(I33, I34, I35, I36, I37) 
  8) f3#(I43, I44, I45, I46, I47) -> f4#(I43, I44, I45, I46, I47) 
  9) f1#(I48, I49, I50, I51, I52) -> f2#(I48, I49, I50, I51, I52) 

We have the following SCCs.
  { 3, 4, 5, 8, 9 }
  { 6, 7 }

DP problem for innermost termination.
P =
  f8#(I28, I29, I30, I31, I32) -> f7#(I28, I29, I30, I31, I32) 
  f7#(I33, I34, I35, I36, I37) -> f8#(I33, I34, I35, I36, I37) 
R = 
  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
  f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8]
  f2(I10, I11, I12, I13, I14) -> f3(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
  f4(I15, I16, I17, I18, I19) -> f1(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
  f4(I23, I24, I25, I26, I27) -> f3(I23, I24, I25, I26, I27) [1 <= I27]
  f8(I28, I29, I30, I31, I32) -> f7(I28, I29, I30, I31, I32) 
  f7(I33, I34, I35, I36, I37) -> f8(I33, I34, I35, I36, I37) 
  f5(I38, I39, I40, I41, I42) -> f6(I38, I39, I40, I41, I42) 
  f3(I43, I44, I45, I46, I47) -> f4(I43, I44, I45, I46, I47) 
  f1(I48, I49, I50, I51, I52) -> f2(I48, I49, I50, I51, I52)