/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 =
  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
  f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  f5#(I8, I9, I10, I11) -> f6#(I8, I9, -1 + I10, -1 + I10) 
  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 + I22 + I23 /\ 0 <= -1 + I23 /\ 0 <= -1 + I22]
  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
  f1#(I28, I29, I30, I31) -> f3#(I28, I32, -2 + I31, 1 + -2 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 + I30 + I31 /\ 0 <= -1 + I31 /\ 0 <= -1 + I30]
R = 
  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
  f5(I8, I9, I10, I11) -> f6(I8, I9, -1 + I10, -1 + I10) 
  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 + I22 + I23 /\ 0 <= -1 + I23 /\ 0 <= -1 + I22]
  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
  f1(I28, I29, I30, I31) -> f3(I28, I32, -2 + I31, 1 + -2 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 + I30 + I31 /\ 0 <= -1 + I31 /\ 0 <= -1 + I30]
  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ I35 + I36 <= 0 /\ 0 <= -1 + I36 /\ 0 <= -1 + I35]
  f1(I37, I38, I39, I40) -> f2(I41, I38, I39, I40) [I41 = I41 /\ I40 <= 0 /\ 0 <= -1 + I39]
  f1(I42, I43, I44, I45) -> f2(I46, I43, I44, I45) [I46 = I46 /\ I44 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 6, 8
  2 -> 6, 8
  3 -> 2
  4 -> 3
  5 -> 3
  6 -> 4, 5
  7 -> 6, 8
  8 -> 7
Where:
  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
  1) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
  2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, -1 + I10, -1 + I10) 
  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
  5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
  6) f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 + I22 + I23 /\ 0 <= -1 + I23 /\ 0 <= -1 + I22]
  7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
  8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, -2 + I31, 1 + -2 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 + I30 + I31 /\ 0 <= -1 + I31 /\ 0 <= -1 + I30]

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7, 8 }

DP problem for innermost termination.
P =
  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  f5#(I8, I9, I10, I11) -> f6#(I8, I9, -1 + I10, -1 + I10) 
  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 + I22 + I23 /\ 0 <= -1 + I23 /\ 0 <= -1 + I22]
  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
  f1#(I28, I29, I30, I31) -> f3#(I28, I32, -2 + I31, 1 + -2 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 + I30 + I31 /\ 0 <= -1 + I31 /\ 0 <= -1 + I30]
R = 
  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
  f5(I8, I9, I10, I11) -> f6(I8, I9, -1 + I10, -1 + I10) 
  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 + I22 + I23 /\ 0 <= -1 + I23 /\ 0 <= -1 + I22]
  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
  f1(I28, I29, I30, I31) -> f3(I28, I32, -2 + I31, 1 + -2 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 + I30 + I31 /\ 0 <= -1 + I31 /\ 0 <= -1 + I30]
  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ I35 + I36 <= 0 /\ 0 <= -1 + I36 /\ 0 <= -1 + I35]
  f1(I37, I38, I39, I40) -> f2(I41, I38, I39, I40) [I41 = I41 /\ I40 <= 0 /\ 0 <= -1 + I39]
  f1(I42, I43, I44, I45) -> f2(I46, I43, I44, I45) [I46 = I46 /\ I44 <= 0]