/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3#(I3, I4, I5) -> f4#(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2#(I9, I10, I11) -> f3#(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1#(I15, I16, I17) -> f2#(0, 0, 0) 
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

The dependency graph for this problem is:
  0 -> 6
  1 -> 1
  2 -> 1
  3 -> 2, 3
  4 -> 3
  5 -> 4, 5
  6 -> 5
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  2) f3#(I3, I4, I5) -> f4#(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  3) f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  4) f2#(I9, I10, I11) -> f3#(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  5) f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  6) f1#(I15, I16, I17) -> f2#(0, 0, 0) 

We have the following SCCs.
  { 5 }
  { 3 }
  { 1 }

DP problem for innermost termination.
P =
  f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

We use the reverse value criterion with the projection function NU:
  NU[f4#(z1,z2,z3)] = 1000 + -1 * z2

This gives the following inequalities:
  I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1  ==>  1000 + -1 * I1 > 1000 + -1 * (I1 + 3)  with  1000 + -1 * I1 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2,z3)] = 1000 + -1 * z2

This gives the following inequalities:
  I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1  ==>  1000 + -1 * I7 > 1000 + -1 * (I7 + 2)  with  1000 + -1 * I7 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = 1000 + -1 * z2

This gives the following inequalities:
  I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1  ==>  1000 + -1 * I13 > 1000 + -1 * (I13 + 1)  with  1000 + -1 * I13 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.