/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 =
  f7#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
  f2#(I7, I8, I9, I10, I11, I12, I13) -> f3#(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
  f2#(I14, I15, I16, I17, I18, I19, I20) -> f4#(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
  f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) 
  f3#(I29, I30, I31, I32, I33, I34, I35) -> f5#(I29, I30, I31, I32, I33, I34, -1 * I34 + I35) [0 <= I35]
  f3#(I36, I37, I38, I39, I40, I41, I42) -> f4#(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
  f1#(I45, I46, I47, I48, I49, I50, I51) -> f2#(I45, I46, I47, I48, I49, I50, I51) 
R = 
  f7(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
  f4(I0, I1, I2, I3, I4, I5, I6) -> f6(rnd1, rnd2, I2, 0, I4, I5, I6) [rnd1 = rnd2 /\ rnd2 = 0]
  f2(I7, I8, I9, I10, I11, I12, I13) -> f3(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
  f2(I14, I15, I16, I17, I18, I19, I20) -> f4(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
  f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) 
  f3(I29, I30, I31, I32, I33, I34, I35) -> f5(I29, I30, I31, I32, I33, I34, -1 * I34 + I35) [0 <= I35]
  f3(I36, I37, I38, I39, I40, I41, I42) -> f4(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
  f1(I45, I46, I47, I48, I49, I50, I51) -> f2(I45, I46, I47, I48, I49, I50, I51) 

The dependency graph for this problem is:
  0 -> 6
  1 -> 4, 5
  2 -> 
  3 -> 4, 5
  4 -> 3
  5 -> 
  6 -> 1, 2
Where:
  0) f7#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
  1) f2#(I7, I8, I9, I10, I11, I12, I13) -> f3#(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
  2) f2#(I14, I15, I16, I17, I18, I19, I20) -> f4#(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
  3) f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) 
  4) f3#(I29, I30, I31, I32, I33, I34, I35) -> f5#(I29, I30, I31, I32, I33, I34, -1 * I34 + I35) [0 <= I35]
  5) f3#(I36, I37, I38, I39, I40, I41, I42) -> f4#(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
  6) f1#(I45, I46, I47, I48, I49, I50, I51) -> f2#(I45, I46, I47, I48, I49, I50, I51) 

We have the following SCCs.
  { 3, 4 }

DP problem for innermost termination.
P =
  f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) 
  f3#(I29, I30, I31, I32, I33, I34, I35) -> f5#(I29, I30, I31, I32, I33, I34, -1 * I34 + I35) [0 <= I35]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
  f4(I0, I1, I2, I3, I4, I5, I6) -> f6(rnd1, rnd2, I2, 0, I4, I5, I6) [rnd1 = rnd2 /\ rnd2 = 0]
  f2(I7, I8, I9, I10, I11, I12, I13) -> f3(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
  f2(I14, I15, I16, I17, I18, I19, I20) -> f4(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
  f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) 
  f3(I29, I30, I31, I32, I33, I34, I35) -> f5(I29, I30, I31, I32, I33, I34, -1 * I34 + I35) [0 <= I35]
  f3(I36, I37, I38, I39, I40, I41, I42) -> f4(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
  f1(I45, I46, I47, I48, I49, I50, I51) -> f2(I45, I46, I47, I48, I49, I50, I51)