/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 =
  f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
  f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1]
  f4#(I5, I6, I7, I8, I9) -> f3#(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9]
  f4#(I10, I11, I12, I13, I14) -> f2#(I10, I11, I12, I13, I14) [I14 <= 0]
  f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, -1 + I19) [1 <= I15]
  f3#(I20, I21, I22, I23, I24) -> f4#(I20, I21, I22, I23, -2 + I24) [I20 <= 0]
  f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2]
  f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31]
  f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0]
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1]
  f4(I5, I6, I7, I8, I9) -> f3(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9]
  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [I14 <= 0]
  f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, -1 + I19) [1 <= I15]
  f3(I20, I21, I22, I23, I24) -> f4(I20, I21, I22, I23, -2 + I24) [I20 <= 0]
  f2(I25, I26, I27, I28, I29) -> f1(I25, rnd2, I27, I28, I29) [rnd2 = rnd2]
  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, 1, I33, I34) [1 <= I31]
  f1(I35, I36, I37, I38, I39) -> f2(I35, I36, 0, I38, I39) [I36 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2, 3
  2 -> 4, 5
  3 -> 6
  4 -> 2, 3
  5 -> 2, 3
  6 -> 7, 8
  7 -> 6
  8 -> 6
Where:
  0) f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
  1) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1]
  2) f4#(I5, I6, I7, I8, I9) -> f3#(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9]
  3) f4#(I10, I11, I12, I13, I14) -> f2#(I10, I11, I12, I13, I14) [I14 <= 0]
  4) f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, -1 + I19) [1 <= I15]
  5) f3#(I20, I21, I22, I23, I24) -> f4#(I20, I21, I22, I23, -2 + I24) [I20 <= 0]
  6) f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2]
  7) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31]
  8) f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0]

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

DP problem for innermost termination.
P =
  f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2]
  f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31]
  f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0]
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1]
  f4(I5, I6, I7, I8, I9) -> f3(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9]
  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [I14 <= 0]
  f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, -1 + I19) [1 <= I15]
  f3(I20, I21, I22, I23, I24) -> f4(I20, I21, I22, I23, -2 + I24) [I20 <= 0]
  f2(I25, I26, I27, I28, I29) -> f1(I25, rnd2, I27, I28, I29) [rnd2 = rnd2]
  f1(I30, I31, I32, I33, I34) -> f2(I30, I31, 1, I33, I34) [1 <= I31]
  f1(I35, I36, I37, I38, I39) -> f2(I35, I36, 0, I38, I39) [I36 <= 0]