/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 =
  f14#(x1, x2) -> f13#(x1, x2) 
  f13#(I0, I1) -> f4#(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  f3#(I2, I3) -> f5#(1 + I2, I3) [I2 <= 5]
  f3#(I4, I5) -> f5#(1 + I4, I5) [6 <= I4]
  f2#(I6, I7) -> f3#(I6, I7) [1 + I7 <= 0]
  f2#(I8, I9) -> f3#(I8, I9) [1 <= I9]
  f2#(I10, I11) -> f6#(I10, I11) [0 <= I11 /\ I11 <= 0]
  f7#(I12, I13) -> f6#(-1 + I12, I13) [3 <= I12]
  f7#(I14, I15) -> f4#(I14, I15) [I14 <= 2]
  f12#(I16, I17) -> f11#(I16, I17) 
  f11#(I18, I19) -> f12#(I18, I19) 
  f10#(I20, I21) -> f11#(I20, I21) 
  f6#(I24, I25) -> f7#(I24, I25) 
  f5#(I26, I27) -> f1#(I26, I27) 
  f4#(I28, I29) -> f5#(I28, I29) 
  f1#(I30, I31) -> f3#(I30, I31) [I30 <= 5]
  f1#(I32, I33) -> f2#(I32, rnd2) [rnd2 = rnd2 /\ 6 <= I32]
R = 
  f14(x1, x2) -> f13(x1, x2) 
  f13(I0, I1) -> f4(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  f3(I2, I3) -> f5(1 + I2, I3) [I2 <= 5]
  f3(I4, I5) -> f5(1 + I4, I5) [6 <= I4]
  f2(I6, I7) -> f3(I6, I7) [1 + I7 <= 0]
  f2(I8, I9) -> f3(I8, I9) [1 <= I9]
  f2(I10, I11) -> f6(I10, I11) [0 <= I11 /\ I11 <= 0]
  f7(I12, I13) -> f6(-1 + I12, I13) [3 <= I12]
  f7(I14, I15) -> f4(I14, I15) [I14 <= 2]
  f12(I16, I17) -> f11(I16, I17) 
  f11(I18, I19) -> f12(I18, I19) 
  f10(I20, I21) -> f11(I20, I21) 
  f8(I22, I23) -> f9(I22, I23) 
  f6(I24, I25) -> f7(I24, I25) 
  f5(I26, I27) -> f1(I26, I27) 
  f4(I28, I29) -> f5(I28, I29) 
  f1(I30, I31) -> f3(I30, I31) [I30 <= 5]
  f1(I32, I33) -> f2(I32, rnd2) [rnd2 = rnd2 /\ 6 <= I32]

The dependency graph for this problem is:
  0 -> 1
  1 -> 14
  2 -> 13
  3 -> 13
  4 -> 2, 3
  5 -> 2, 3
  6 -> 12
  7 -> 12
  8 -> 14
  9 -> 10
  10 -> 9
  11 -> 10
  12 -> 7, 8
  13 -> 15, 16
  14 -> 13
  15 -> 2
  16 -> 4, 5, 6
Where:
  0) f14#(x1, x2) -> f13#(x1, x2) 
  1) f13#(I0, I1) -> f4#(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  2) f3#(I2, I3) -> f5#(1 + I2, I3) [I2 <= 5]
  3) f3#(I4, I5) -> f5#(1 + I4, I5) [6 <= I4]
  4) f2#(I6, I7) -> f3#(I6, I7) [1 + I7 <= 0]
  5) f2#(I8, I9) -> f3#(I8, I9) [1 <= I9]
  6) f2#(I10, I11) -> f6#(I10, I11) [0 <= I11 /\ I11 <= 0]
  7) f7#(I12, I13) -> f6#(-1 + I12, I13) [3 <= I12]
  8) f7#(I14, I15) -> f4#(I14, I15) [I14 <= 2]
  9) f12#(I16, I17) -> f11#(I16, I17) 
  10) f11#(I18, I19) -> f12#(I18, I19) 
  11) f10#(I20, I21) -> f11#(I20, I21) 
  12) f6#(I24, I25) -> f7#(I24, I25) 
  13) f5#(I26, I27) -> f1#(I26, I27) 
  14) f4#(I28, I29) -> f5#(I28, I29) 
  15) f1#(I30, I31) -> f3#(I30, I31) [I30 <= 5]
  16) f1#(I32, I33) -> f2#(I32, rnd2) [rnd2 = rnd2 /\ 6 <= I32]

We have the following SCCs.
  { 9, 10 }
  { 2, 3, 4, 5, 6, 7, 8, 12, 13, 14, 15, 16 }

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f5#(1 + I2, I3) [I2 <= 5]
  f3#(I4, I5) -> f5#(1 + I4, I5) [6 <= I4]
  f2#(I6, I7) -> f3#(I6, I7) [1 + I7 <= 0]
  f2#(I8, I9) -> f3#(I8, I9) [1 <= I9]
  f2#(I10, I11) -> f6#(I10, I11) [0 <= I11 /\ I11 <= 0]
  f7#(I12, I13) -> f6#(-1 + I12, I13) [3 <= I12]
  f7#(I14, I15) -> f4#(I14, I15) [I14 <= 2]
  f6#(I24, I25) -> f7#(I24, I25) 
  f5#(I26, I27) -> f1#(I26, I27) 
  f4#(I28, I29) -> f5#(I28, I29) 
  f1#(I30, I31) -> f3#(I30, I31) [I30 <= 5]
  f1#(I32, I33) -> f2#(I32, rnd2) [rnd2 = rnd2 /\ 6 <= I32]
R = 
  f14(x1, x2) -> f13(x1, x2) 
  f13(I0, I1) -> f4(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  f3(I2, I3) -> f5(1 + I2, I3) [I2 <= 5]
  f3(I4, I5) -> f5(1 + I4, I5) [6 <= I4]
  f2(I6, I7) -> f3(I6, I7) [1 + I7 <= 0]
  f2(I8, I9) -> f3(I8, I9) [1 <= I9]
  f2(I10, I11) -> f6(I10, I11) [0 <= I11 /\ I11 <= 0]
  f7(I12, I13) -> f6(-1 + I12, I13) [3 <= I12]
  f7(I14, I15) -> f4(I14, I15) [I14 <= 2]
  f12(I16, I17) -> f11(I16, I17) 
  f11(I18, I19) -> f12(I18, I19) 
  f10(I20, I21) -> f11(I20, I21) 
  f8(I22, I23) -> f9(I22, I23) 
  f6(I24, I25) -> f7(I24, I25) 
  f5(I26, I27) -> f1(I26, I27) 
  f4(I28, I29) -> f5(I28, I29) 
  f1(I30, I31) -> f3(I30, I31) [I30 <= 5]
  f1(I32, I33) -> f2(I32, rnd2) [rnd2 = rnd2 /\ 6 <= I32]