/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 =
  f11#(x1, x2, x3) -> f10#(x1, x2, x3) 
  f10#(I0, I1, I2) -> f8#(0, I1, I2) 
  f3#(I3, I4, I5) -> f8#(I3, I4, I5) 
  f9#(I6, I7, I8) -> f6#(0, I7, I8) [1 <= I6 /\ I6 <= 1]
  f9#(I9, I10, I11) -> f6#(I9, I10, I11) [1 + I9 <= 1]
  f9#(I12, I13, I14) -> f6#(I12, I13, I14) [2 <= I12]
  f2#(I15, I16, I17) -> f9#(I15, I16, I17) 
  f8#(I18, I19, I20) -> f5#(I18, I19, rnd3) [rnd3 = rnd3]
  f5#(I24, I25, I26) -> f4#(I24, I25, I26) [1 + I26 <= 0]
  f5#(I27, I28, I29) -> f4#(I27, I28, I29) [1 <= I29]
  f5#(I30, I31, I32) -> f2#(I30, I31, I32) [0 <= I32 /\ I32 <= 0]
  f4#(I33, I34, I35) -> f1#(I33, rnd2, I35) [rnd2 = rnd2]
  f1#(I36, I37, I38) -> f3#(I36, I37, I38) [1 + I37 <= 0]
  f1#(I39, I40, I41) -> f3#(I39, I40, I41) [1 <= I40]
  f1#(I42, I43, I44) -> f2#(1, I43, I44) [0 <= I43 /\ I43 <= 0]
R = 
  f11(x1, x2, x3) -> f10(x1, x2, x3) 
  f10(I0, I1, I2) -> f8(0, I1, I2) 
  f3(I3, I4, I5) -> f8(I3, I4, I5) 
  f9(I6, I7, I8) -> f6(0, I7, I8) [1 <= I6 /\ I6 <= 1]
  f9(I9, I10, I11) -> f6(I9, I10, I11) [1 + I9 <= 1]
  f9(I12, I13, I14) -> f6(I12, I13, I14) [2 <= I12]
  f2(I15, I16, I17) -> f9(I15, I16, I17) 
  f8(I18, I19, I20) -> f5(I18, I19, rnd3) [rnd3 = rnd3]
  f6(I21, I22, I23) -> f7(I21, I22, I23) 
  f5(I24, I25, I26) -> f4(I24, I25, I26) [1 + I26 <= 0]
  f5(I27, I28, I29) -> f4(I27, I28, I29) [1 <= I29]
  f5(I30, I31, I32) -> f2(I30, I31, I32) [0 <= I32 /\ I32 <= 0]
  f4(I33, I34, I35) -> f1(I33, rnd2, I35) [rnd2 = rnd2]
  f1(I36, I37, I38) -> f3(I36, I37, I38) [1 + I37 <= 0]
  f1(I39, I40, I41) -> f3(I39, I40, I41) [1 <= I40]
  f1(I42, I43, I44) -> f2(1, I43, I44) [0 <= I43 /\ I43 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 7
  2 -> 7
  3 -> 
  4 -> 
  5 -> 
  6 -> 3, 4, 5
  7 -> 8, 9, 10
  8 -> 11
  9 -> 11
  10 -> 6
  11 -> 12, 13, 14
  12 -> 2
  13 -> 2
  14 -> 6
Where:
  0) f11#(x1, x2, x3) -> f10#(x1, x2, x3) 
  1) f10#(I0, I1, I2) -> f8#(0, I1, I2) 
  2) f3#(I3, I4, I5) -> f8#(I3, I4, I5) 
  3) f9#(I6, I7, I8) -> f6#(0, I7, I8) [1 <= I6 /\ I6 <= 1]
  4) f9#(I9, I10, I11) -> f6#(I9, I10, I11) [1 + I9 <= 1]
  5) f9#(I12, I13, I14) -> f6#(I12, I13, I14) [2 <= I12]
  6) f2#(I15, I16, I17) -> f9#(I15, I16, I17) 
  7) f8#(I18, I19, I20) -> f5#(I18, I19, rnd3) [rnd3 = rnd3]
  8) f5#(I24, I25, I26) -> f4#(I24, I25, I26) [1 + I26 <= 0]
  9) f5#(I27, I28, I29) -> f4#(I27, I28, I29) [1 <= I29]
  10) f5#(I30, I31, I32) -> f2#(I30, I31, I32) [0 <= I32 /\ I32 <= 0]
  11) f4#(I33, I34, I35) -> f1#(I33, rnd2, I35) [rnd2 = rnd2]
  12) f1#(I36, I37, I38) -> f3#(I36, I37, I38) [1 + I37 <= 0]
  13) f1#(I39, I40, I41) -> f3#(I39, I40, I41) [1 <= I40]
  14) f1#(I42, I43, I44) -> f2#(1, I43, I44) [0 <= I43 /\ I43 <= 0]

We have the following SCCs.
  { 2, 7, 8, 9, 11, 12, 13 }

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f8#(I3, I4, I5) 
  f8#(I18, I19, I20) -> f5#(I18, I19, rnd3) [rnd3 = rnd3]
  f5#(I24, I25, I26) -> f4#(I24, I25, I26) [1 + I26 <= 0]
  f5#(I27, I28, I29) -> f4#(I27, I28, I29) [1 <= I29]
  f4#(I33, I34, I35) -> f1#(I33, rnd2, I35) [rnd2 = rnd2]
  f1#(I36, I37, I38) -> f3#(I36, I37, I38) [1 + I37 <= 0]
  f1#(I39, I40, I41) -> f3#(I39, I40, I41) [1 <= I40]
R = 
  f11(x1, x2, x3) -> f10(x1, x2, x3) 
  f10(I0, I1, I2) -> f8(0, I1, I2) 
  f3(I3, I4, I5) -> f8(I3, I4, I5) 
  f9(I6, I7, I8) -> f6(0, I7, I8) [1 <= I6 /\ I6 <= 1]
  f9(I9, I10, I11) -> f6(I9, I10, I11) [1 + I9 <= 1]
  f9(I12, I13, I14) -> f6(I12, I13, I14) [2 <= I12]
  f2(I15, I16, I17) -> f9(I15, I16, I17) 
  f8(I18, I19, I20) -> f5(I18, I19, rnd3) [rnd3 = rnd3]
  f6(I21, I22, I23) -> f7(I21, I22, I23) 
  f5(I24, I25, I26) -> f4(I24, I25, I26) [1 + I26 <= 0]
  f5(I27, I28, I29) -> f4(I27, I28, I29) [1 <= I29]
  f5(I30, I31, I32) -> f2(I30, I31, I32) [0 <= I32 /\ I32 <= 0]
  f4(I33, I34, I35) -> f1(I33, rnd2, I35) [rnd2 = rnd2]
  f1(I36, I37, I38) -> f3(I36, I37, I38) [1 + I37 <= 0]
  f1(I39, I40, I41) -> f3(I39, I40, I41) [1 <= I40]
  f1(I42, I43, I44) -> f2(1, I43, I44) [0 <= I43 /\ I43 <= 0]