/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 =
  f7#(x1, x2) -> f6#(x1, x2) 
  f6#(I0, I1) -> f2#(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
  f5#(I2, I3) -> f1#(I2, rnd2) [rnd2 = rnd2]
  f4#(I4, I5) -> f5#(I4, I5) [5 <= I4]
  f4#(I6, I7) -> f5#(I6, I7) [1 + I6 <= 4]
  f3#(I8, I9) -> f4#(I8, I9) [3 <= I8]
  f3#(I10, I11) -> f4#(I10, I11) [1 + I10 <= 2]
  f2#(I12, I13) -> f3#(I12, I13) [2 <= I12]
  f2#(I14, I15) -> f3#(I14, I15) [1 + I14 <= 1]
  f1#(I16, I17) -> f2#(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
  f1#(I18, I19) -> f2#(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]
R = 
  f7(x1, x2) -> f6(x1, x2) 
  f6(I0, I1) -> f2(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
  f5(I2, I3) -> f1(I2, rnd2) [rnd2 = rnd2]
  f4(I4, I5) -> f5(I4, I5) [5 <= I4]
  f4(I6, I7) -> f5(I6, I7) [1 + I6 <= 4]
  f3(I8, I9) -> f4(I8, I9) [3 <= I8]
  f3(I10, I11) -> f4(I10, I11) [1 + I10 <= 2]
  f2(I12, I13) -> f3(I12, I13) [2 <= I12]
  f2(I14, I15) -> f3(I14, I15) [1 + I14 <= 1]
  f1(I16, I17) -> f2(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
  f1(I18, I19) -> f2(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]

The dependency graph for this problem is:
  0 -> 1
  1 -> 7
  2 -> 9, 10
  3 -> 2
  4 -> 2
  5 -> 3, 4
  6 -> 4
  7 -> 5
  8 -> 6
  9 -> 7, 8
  10 -> 7, 8
Where:
  0) f7#(x1, x2) -> f6#(x1, x2) 
  1) f6#(I0, I1) -> f2#(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
  2) f5#(I2, I3) -> f1#(I2, rnd2) [rnd2 = rnd2]
  3) f4#(I4, I5) -> f5#(I4, I5) [5 <= I4]
  4) f4#(I6, I7) -> f5#(I6, I7) [1 + I6 <= 4]
  5) f3#(I8, I9) -> f4#(I8, I9) [3 <= I8]
  6) f3#(I10, I11) -> f4#(I10, I11) [1 + I10 <= 2]
  7) f2#(I12, I13) -> f3#(I12, I13) [2 <= I12]
  8) f2#(I14, I15) -> f3#(I14, I15) [1 + I14 <= 1]
  9) f1#(I16, I17) -> f2#(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
  10) f1#(I18, I19) -> f2#(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7, 8, 9, 10 }

DP problem for innermost termination.
P =
  f5#(I2, I3) -> f1#(I2, rnd2) [rnd2 = rnd2]
  f4#(I4, I5) -> f5#(I4, I5) [5 <= I4]
  f4#(I6, I7) -> f5#(I6, I7) [1 + I6 <= 4]
  f3#(I8, I9) -> f4#(I8, I9) [3 <= I8]
  f3#(I10, I11) -> f4#(I10, I11) [1 + I10 <= 2]
  f2#(I12, I13) -> f3#(I12, I13) [2 <= I12]
  f2#(I14, I15) -> f3#(I14, I15) [1 + I14 <= 1]
  f1#(I16, I17) -> f2#(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
  f1#(I18, I19) -> f2#(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]
R = 
  f7(x1, x2) -> f6(x1, x2) 
  f6(I0, I1) -> f2(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
  f5(I2, I3) -> f1(I2, rnd2) [rnd2 = rnd2]
  f4(I4, I5) -> f5(I4, I5) [5 <= I4]
  f4(I6, I7) -> f5(I6, I7) [1 + I6 <= 4]
  f3(I8, I9) -> f4(I8, I9) [3 <= I8]
  f3(I10, I11) -> f4(I10, I11) [1 + I10 <= 2]
  f2(I12, I13) -> f3(I12, I13) [2 <= I12]
  f2(I14, I15) -> f3(I14, I15) [1 + I14 <= 1]
  f1(I16, I17) -> f2(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
  f1(I18, I19) -> f2(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]