/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 =
  f15#(x1, x2) -> f14#(x1, x2) 
  f14#(I0, I1) -> f4#(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  f3#(I2, I3) -> f13#(I2, I3) [1 + I3 <= 0]
  f3#(I4, I5) -> f13#(I4, I5) [1 <= I5]
  f3#(I6, I7) -> f2#(I6, I7) [0 <= I7 /\ I7 <= 0]
  f13#(I8, I9) -> f5#(1 + I8, I9) [I8 <= 5]
  f13#(I10, I11) -> f5#(1 + I10, I11) [6 <= I10]
  f2#(I12, I13) -> f6#(I12, I13) 
  f12#(I14, I15) -> f11#(I14, I15) 
  f11#(I16, I17) -> f12#(I16, I17) 
  f7#(I18, I19) -> f6#(-1 + I18, I19) [3 <= I18]
  f7#(I20, I21) -> f4#(I20, I21) [I20 <= 2]
  f10#(I22, I23) -> f11#(I22, I23) 
  f6#(I26, I27) -> f7#(I26, I27) 
  f5#(I28, I29) -> f1#(I28, I29) 
  f4#(I30, I31) -> f5#(I30, I31) 
  f1#(I32, I33) -> f3#(I32, rnd2) [rnd2 = rnd2 /\ I32 <= 5]
  f1#(I34, I35) -> f2#(I34, I35) [6 <= I34]
R = 
  f15(x1, x2) -> f14(x1, x2) 
  f14(I0, I1) -> f4(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  f3(I2, I3) -> f13(I2, I3) [1 + I3 <= 0]
  f3(I4, I5) -> f13(I4, I5) [1 <= I5]
  f3(I6, I7) -> f2(I6, I7) [0 <= I7 /\ I7 <= 0]
  f13(I8, I9) -> f5(1 + I8, I9) [I8 <= 5]
  f13(I10, I11) -> f5(1 + I10, I11) [6 <= I10]
  f2(I12, I13) -> f6(I12, I13) 
  f12(I14, I15) -> f11(I14, I15) 
  f11(I16, I17) -> f12(I16, I17) 
  f7(I18, I19) -> f6(-1 + I18, I19) [3 <= I18]
  f7(I20, I21) -> f4(I20, I21) [I20 <= 2]
  f10(I22, I23) -> f11(I22, I23) 
  f8(I24, I25) -> f9(I24, I25) 
  f6(I26, I27) -> f7(I26, I27) 
  f5(I28, I29) -> f1(I28, I29) 
  f4(I30, I31) -> f5(I30, I31) 
  f1(I32, I33) -> f3(I32, rnd2) [rnd2 = rnd2 /\ I32 <= 5]
  f1(I34, I35) -> f2(I34, I35) [6 <= I34]

The dependency graph for this problem is:
  0 -> 1
  1 -> 15
  2 -> 5, 6
  3 -> 5, 6
  4 -> 7
  5 -> 14
  6 -> 14
  7 -> 13
  8 -> 9
  9 -> 8
  10 -> 13
  11 -> 15
  12 -> 9
  13 -> 10, 11
  14 -> 16, 17
  15 -> 14
  16 -> 2, 3, 4
  17 -> 7
Where:
  0) f15#(x1, x2) -> f14#(x1, x2) 
  1) f14#(I0, I1) -> f4#(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  2) f3#(I2, I3) -> f13#(I2, I3) [1 + I3 <= 0]
  3) f3#(I4, I5) -> f13#(I4, I5) [1 <= I5]
  4) f3#(I6, I7) -> f2#(I6, I7) [0 <= I7 /\ I7 <= 0]
  5) f13#(I8, I9) -> f5#(1 + I8, I9) [I8 <= 5]
  6) f13#(I10, I11) -> f5#(1 + I10, I11) [6 <= I10]
  7) f2#(I12, I13) -> f6#(I12, I13) 
  8) f12#(I14, I15) -> f11#(I14, I15) 
  9) f11#(I16, I17) -> f12#(I16, I17) 
  10) f7#(I18, I19) -> f6#(-1 + I18, I19) [3 <= I18]
  11) f7#(I20, I21) -> f4#(I20, I21) [I20 <= 2]
  12) f10#(I22, I23) -> f11#(I22, I23) 
  13) f6#(I26, I27) -> f7#(I26, I27) 
  14) f5#(I28, I29) -> f1#(I28, I29) 
  15) f4#(I30, I31) -> f5#(I30, I31) 
  16) f1#(I32, I33) -> f3#(I32, rnd2) [rnd2 = rnd2 /\ I32 <= 5]
  17) f1#(I34, I35) -> f2#(I34, I35) [6 <= I34]

We have the following SCCs.
  { 8, 9 }
  { 2, 3, 4, 5, 6, 7, 10, 11, 13, 14, 15, 16, 17 }

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f13#(I2, I3) [1 + I3 <= 0]
  f3#(I4, I5) -> f13#(I4, I5) [1 <= I5]
  f3#(I6, I7) -> f2#(I6, I7) [0 <= I7 /\ I7 <= 0]
  f13#(I8, I9) -> f5#(1 + I8, I9) [I8 <= 5]
  f13#(I10, I11) -> f5#(1 + I10, I11) [6 <= I10]
  f2#(I12, I13) -> f6#(I12, I13) 
  f7#(I18, I19) -> f6#(-1 + I18, I19) [3 <= I18]
  f7#(I20, I21) -> f4#(I20, I21) [I20 <= 2]
  f6#(I26, I27) -> f7#(I26, I27) 
  f5#(I28, I29) -> f1#(I28, I29) 
  f4#(I30, I31) -> f5#(I30, I31) 
  f1#(I32, I33) -> f3#(I32, rnd2) [rnd2 = rnd2 /\ I32 <= 5]
  f1#(I34, I35) -> f2#(I34, I35) [6 <= I34]
R = 
  f15(x1, x2) -> f14(x1, x2) 
  f14(I0, I1) -> f4(rnd1, I1) [y1 = y1 /\ rnd1 = rnd1]
  f3(I2, I3) -> f13(I2, I3) [1 + I3 <= 0]
  f3(I4, I5) -> f13(I4, I5) [1 <= I5]
  f3(I6, I7) -> f2(I6, I7) [0 <= I7 /\ I7 <= 0]
  f13(I8, I9) -> f5(1 + I8, I9) [I8 <= 5]
  f13(I10, I11) -> f5(1 + I10, I11) [6 <= I10]
  f2(I12, I13) -> f6(I12, I13) 
  f12(I14, I15) -> f11(I14, I15) 
  f11(I16, I17) -> f12(I16, I17) 
  f7(I18, I19) -> f6(-1 + I18, I19) [3 <= I18]
  f7(I20, I21) -> f4(I20, I21) [I20 <= 2]
  f10(I22, I23) -> f11(I22, I23) 
  f8(I24, I25) -> f9(I24, I25) 
  f6(I26, I27) -> f7(I26, I27) 
  f5(I28, I29) -> f1(I28, I29) 
  f4(I30, I31) -> f5(I30, I31) 
  f1(I32, I33) -> f3(I32, rnd2) [rnd2 = rnd2 /\ I32 <= 5]
  f1(I34, I35) -> f2(I34, I35) [6 <= I34]