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


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
  f4#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  f6#(I8, I9, I10, I11) -> f4#(I8, 0, I10, I11) 
  f3#(I17, I18, I19, I20) -> f4#(I17, 0, I19, -1 + I20) 
  f2#(I21, I22, I23, I24) -> f3#(I21, I22, I23, I24) [1 <= I22]
  f2#(I25, I26, I27, I28) -> f3#(I25, I26, I27, I28) [1 + I26 <= 0]
  f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) [0 <= -1 - I31 + I32]
R = 
  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f5(rnd1, I1, I2, I3) [rnd1 = rnd1 /\ -1 * I2 + I3 <= 0]
  f4(I4, I5, I6, I7) -> f1(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  f6(I8, I9, I10, I11) -> f4(I8, 0, I10, I11) 
  f1(I12, I13, I14, I15) -> f5(I16, I13, I14, I15) [I16 = I16 /\ -1 * I14 + I15 <= 0]
  f3(I17, I18, I19, I20) -> f4(I17, 0, I19, -1 + I20) 
  f2(I21, I22, I23, I24) -> f3(I21, I22, I23, I24) [1 <= I22]
  f2(I25, I26, I27, I28) -> f3(I25, I26, I27, I28) [1 + I26 <= 0]
  f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) [0 <= -1 - I31 + I32]

The dependency graph for this problem is:
  0 -> 2
  1 -> 6
  2 -> 1
  3 -> 1
  4 -> 3
  5 -> 3
  6 -> 4, 5
Where:
  0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
  1) f4#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  2) f6#(I8, I9, I10, I11) -> f4#(I8, 0, I10, I11) 
  3) f3#(I17, I18, I19, I20) -> f4#(I17, 0, I19, -1 + I20) 
  4) f2#(I21, I22, I23, I24) -> f3#(I21, I22, I23, I24) [1 <= I22]
  5) f2#(I25, I26, I27, I28) -> f3#(I25, I26, I27, I28) [1 + I26 <= 0]
  6) f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) [0 <= -1 - I31 + I32]

We have the following SCCs.
  { 1, 3, 4, 5, 6 }

DP problem for innermost termination.
P =
  f4#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  f3#(I17, I18, I19, I20) -> f4#(I17, 0, I19, -1 + I20) 
  f2#(I21, I22, I23, I24) -> f3#(I21, I22, I23, I24) [1 <= I22]
  f2#(I25, I26, I27, I28) -> f3#(I25, I26, I27, I28) [1 + I26 <= 0]
  f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) [0 <= -1 - I31 + I32]
R = 
  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f5(rnd1, I1, I2, I3) [rnd1 = rnd1 /\ -1 * I2 + I3 <= 0]
  f4(I4, I5, I6, I7) -> f1(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  f6(I8, I9, I10, I11) -> f4(I8, 0, I10, I11) 
  f1(I12, I13, I14, I15) -> f5(I16, I13, I14, I15) [I16 = I16 /\ -1 * I14 + I15 <= 0]
  f3(I17, I18, I19, I20) -> f4(I17, 0, I19, -1 + I20) 
  f2(I21, I22, I23, I24) -> f3(I21, I22, I23, I24) [1 <= I22]
  f2(I25, I26, I27, I28) -> f3(I25, I26, I27, I28) [1 + I26 <= 0]
  f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) [0 <= -1 - I31 + I32]

We use the extended value criterion with the projection function NU:
  NU[f2#(x0,x1,x2,x3)] = -x2 + x3 - 2
  NU[f3#(x0,x1,x2,x3)] = -x2 + x3 - 2
  NU[f1#(x0,x1,x2,x3)] = -x2 + x3 - 1
  NU[f4#(x0,x1,x2,x3)] = x1 - x2 + x3 - 2

This gives the following inequalities:
  0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7  ==>  I5 - I6 + I7 - 2 >= -(1 + I6) + I7 - 1
    ==>  -I19 + I20 - 2 >= 0 - I19 + (-1 + I20) - 2
  1 <= I22  ==>  -I23 + I24 - 2 >= -I23 + I24 - 2
  1 + I26 <= 0  ==>  -I27 + I28 - 2 >= -I27 + I28 - 2
  0 <= -1 - I31 + I32  ==>  -I31 + I32 - 1 > -I31 + I32 - 2  with  -I31 + I32 - 1 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  f3#(I17, I18, I19, I20) -> f4#(I17, 0, I19, -1 + I20) 
  f2#(I21, I22, I23, I24) -> f3#(I21, I22, I23, I24) [1 <= I22]
  f2#(I25, I26, I27, I28) -> f3#(I25, I26, I27, I28) [1 + I26 <= 0]
R = 
  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f5(rnd1, I1, I2, I3) [rnd1 = rnd1 /\ -1 * I2 + I3 <= 0]
  f4(I4, I5, I6, I7) -> f1(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  f6(I8, I9, I10, I11) -> f4(I8, 0, I10, I11) 
  f1(I12, I13, I14, I15) -> f5(I16, I13, I14, I15) [I16 = I16 /\ -1 * I14 + I15 <= 0]
  f3(I17, I18, I19, I20) -> f4(I17, 0, I19, -1 + I20) 
  f2(I21, I22, I23, I24) -> f3(I21, I22, I23, I24) [1 <= I22]
  f2(I25, I26, I27, I28) -> f3(I25, I26, I27, I28) [1 + I26 <= 0]
  f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) [0 <= -1 - I31 + I32]

The dependency graph for this problem is:
  1 -> 
  3 -> 1
  4 -> 3
  5 -> 3
Where:
  1) f4#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= I5 /\ I5 <= 0 /\ 0 <= -1 - I6 + I7]
  3) f3#(I17, I18, I19, I20) -> f4#(I17, 0, I19, -1 + I20) 
  4) f2#(I21, I22, I23, I24) -> f3#(I21, I22, I23, I24) [1 <= I22]
  5) f2#(I25, I26, I27, I28) -> f3#(I25, I26, I27, I28) [1 + I26 <= 0]

We have the following SCCs.