/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 =
  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2 + 4, I3) [6 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ -1 <= I3 - 1 /\ I2 + 1 <= I3 - 1 /\ I2 + 2 <= I3 - 1]
  f4#(I6, I7, I8, I9) -> f4#(I10, I11, I12, I13) [-1 <= I10 - 1 /\ 0 <= I6 - 1]
  f3#(I14, I15, I16, I17) -> f3#(I18, I19, I20, I21) [-1 <= I18 - 1 /\ 0 <= I14 - 1]
  f3#(I22, I23, I24, I25) -> f4#(I26, I27, I28, I29) [-1 <= I26 - 1 /\ -1 <= I22 - 1]
  f2#(I30, I31, I32, I33) -> f3#(I34, I35, I36, I37) [-1 <= I34 - 1 /\ 0 <= I31 - 1 /\ I33 <= I32 /\ 0 <= I30 - 1]
  f1#(I38, I39, I40, I41) -> f2#(I42, I43, 1, I39) [1 <= I43 - 1 /\ 0 <= I42 - 1 /\ 0 <= I38 - 1 /\ I43 - 1 <= I38 /\ -1 <= I39 - 1 /\ I42 <= I38]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f2(I0, I1, I2, I3) -> f2(I4, I5, I2 + 4, I3) [6 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ -1 <= I3 - 1 /\ I2 + 1 <= I3 - 1 /\ I2 + 2 <= I3 - 1]
  f4(I6, I7, I8, I9) -> f4(I10, I11, I12, I13) [-1 <= I10 - 1 /\ 0 <= I6 - 1]
  f3(I14, I15, I16, I17) -> f3(I18, I19, I20, I21) [-1 <= I18 - 1 /\ 0 <= I14 - 1]
  f3(I22, I23, I24, I25) -> f4(I26, I27, I28, I29) [-1 <= I26 - 1 /\ -1 <= I22 - 1]
  f2(I30, I31, I32, I33) -> f3(I34, I35, I36, I37) [-1 <= I34 - 1 /\ 0 <= I31 - 1 /\ I33 <= I32 /\ 0 <= I30 - 1]
  f1(I38, I39, I40, I41) -> f2(I42, I43, 1, I39) [1 <= I43 - 1 /\ 0 <= I42 - 1 /\ 0 <= I38 - 1 /\ I43 - 1 <= I38 /\ -1 <= I39 - 1 /\ I42 <= I38]

The dependency graph for this problem is:
  0 -> 6
  1 -> 1, 5
  2 -> 2
  3 -> 3, 4
  4 -> 2
  5 -> 3, 4
  6 -> 1, 5
Where:
  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  1) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2 + 4, I3) [6 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ -1 <= I3 - 1 /\ I2 + 1 <= I3 - 1 /\ I2 + 2 <= I3 - 1]
  2) f4#(I6, I7, I8, I9) -> f4#(I10, I11, I12, I13) [-1 <= I10 - 1 /\ 0 <= I6 - 1]
  3) f3#(I14, I15, I16, I17) -> f3#(I18, I19, I20, I21) [-1 <= I18 - 1 /\ 0 <= I14 - 1]
  4) f3#(I22, I23, I24, I25) -> f4#(I26, I27, I28, I29) [-1 <= I26 - 1 /\ -1 <= I22 - 1]
  5) f2#(I30, I31, I32, I33) -> f3#(I34, I35, I36, I37) [-1 <= I34 - 1 /\ 0 <= I31 - 1 /\ I33 <= I32 /\ 0 <= I30 - 1]
  6) f1#(I38, I39, I40, I41) -> f2#(I42, I43, 1, I39) [1 <= I43 - 1 /\ 0 <= I42 - 1 /\ 0 <= I38 - 1 /\ I43 - 1 <= I38 /\ -1 <= I39 - 1 /\ I42 <= I38]

We have the following SCCs.
  { 1 }
  { 3 }
  { 2 }

DP problem for innermost termination.
P =
  f4#(I6, I7, I8, I9) -> f4#(I10, I11, I12, I13) [-1 <= I10 - 1 /\ 0 <= I6 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f2(I0, I1, I2, I3) -> f2(I4, I5, I2 + 4, I3) [6 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ -1 <= I3 - 1 /\ I2 + 1 <= I3 - 1 /\ I2 + 2 <= I3 - 1]
  f4(I6, I7, I8, I9) -> f4(I10, I11, I12, I13) [-1 <= I10 - 1 /\ 0 <= I6 - 1]
  f3(I14, I15, I16, I17) -> f3(I18, I19, I20, I21) [-1 <= I18 - 1 /\ 0 <= I14 - 1]
  f3(I22, I23, I24, I25) -> f4(I26, I27, I28, I29) [-1 <= I26 - 1 /\ -1 <= I22 - 1]
  f2(I30, I31, I32, I33) -> f3(I34, I35, I36, I37) [-1 <= I34 - 1 /\ 0 <= I31 - 1 /\ I33 <= I32 /\ 0 <= I30 - 1]
  f1(I38, I39, I40, I41) -> f2(I42, I43, 1, I39) [1 <= I43 - 1 /\ 0 <= I42 - 1 /\ 0 <= I38 - 1 /\ I43 - 1 <= I38 /\ -1 <= I39 - 1 /\ I42 <= I38]