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


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YES

DP problem for innermost termination.
P =
  f8#(x1, x2, x3) -> f1#(x1, x2, x3) 
  f7#(I0, I1, I2) -> f2#(I0, I1, I2) 
  f2#(I3, I4, I5) -> f7#(I3, 1 + I4, I5) [0 <= -1 - I4 + I5]
  f6#(I6, I7, I8) -> f2#(I6, I7, I8) 
  f2#(I9, I10, I11) -> f6#(I9, 1 + I10, I11) [I11 <= I10 /\ I10 <= I11 /\ -1 * I10 + I11 <= 0 /\ -1 * I10 + I11 <= 0]
  f3#(I15, I16, I17) -> f4#(I15, I16, I17) [1 + I17 <= I16]
  f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I19 <= I20]
  f2#(I21, I22, I23) -> f3#(I21, I22, I23) [-1 * I22 + I23 <= 0 /\ -1 * I22 + I23 <= 0]
  f1#(I24, I25, I26) -> f2#(I24, I25, I26) 
R = 
  f8(x1, x2, x3) -> f1(x1, x2, x3) 
  f7(I0, I1, I2) -> f2(I0, I1, I2) 
  f2(I3, I4, I5) -> f7(I3, 1 + I4, I5) [0 <= -1 - I4 + I5]
  f6(I6, I7, I8) -> f2(I6, I7, I8) 
  f2(I9, I10, I11) -> f6(I9, 1 + I10, I11) [I11 <= I10 /\ I10 <= I11 /\ -1 * I10 + I11 <= 0 /\ -1 * I10 + I11 <= 0]
  f4(I12, I13, I14) -> f5(rnd1, I13, I14) [rnd1 = rnd1]
  f3(I15, I16, I17) -> f4(I15, I16, I17) [1 + I17 <= I16]
  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I19 <= I20]
  f2(I21, I22, I23) -> f3(I21, I22, I23) [-1 * I22 + I23 <= 0 /\ -1 * I22 + I23 <= 0]
  f1(I24, I25, I26) -> f2(I24, I25, I26) 

The dependency graph for this problem is:
  0 -> 8
  1 -> 2, 4, 7
  2 -> 1
  3 -> 2, 4, 7
  4 -> 3
  5 -> 
  6 -> 
  7 -> 5
  8 -> 2, 4, 7
Where:
  0) f8#(x1, x2, x3) -> f1#(x1, x2, x3) 
  1) f7#(I0, I1, I2) -> f2#(I0, I1, I2) 
  2) f2#(I3, I4, I5) -> f7#(I3, 1 + I4, I5) [0 <= -1 - I4 + I5]
  3) f6#(I6, I7, I8) -> f2#(I6, I7, I8) 
  4) f2#(I9, I10, I11) -> f6#(I9, 1 + I10, I11) [I11 <= I10 /\ I10 <= I11 /\ -1 * I10 + I11 <= 0 /\ -1 * I10 + I11 <= 0]
  5) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [1 + I17 <= I16]
  6) f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I19 <= I20]
  7) f2#(I21, I22, I23) -> f3#(I21, I22, I23) [-1 * I22 + I23 <= 0 /\ -1 * I22 + I23 <= 0]
  8) f1#(I24, I25, I26) -> f2#(I24, I25, I26) 

We have the following SCCs.
  { 1, 2, 3, 4 }

DP problem for innermost termination.
P =
  f7#(I0, I1, I2) -> f2#(I0, I1, I2) 
  f2#(I3, I4, I5) -> f7#(I3, 1 + I4, I5) [0 <= -1 - I4 + I5]
  f6#(I6, I7, I8) -> f2#(I6, I7, I8) 
  f2#(I9, I10, I11) -> f6#(I9, 1 + I10, I11) [I11 <= I10 /\ I10 <= I11 /\ -1 * I10 + I11 <= 0 /\ -1 * I10 + I11 <= 0]
R = 
  f8(x1, x2, x3) -> f1(x1, x2, x3) 
  f7(I0, I1, I2) -> f2(I0, I1, I2) 
  f2(I3, I4, I5) -> f7(I3, 1 + I4, I5) [0 <= -1 - I4 + I5]
  f6(I6, I7, I8) -> f2(I6, I7, I8) 
  f2(I9, I10, I11) -> f6(I9, 1 + I10, I11) [I11 <= I10 /\ I10 <= I11 /\ -1 * I10 + I11 <= 0 /\ -1 * I10 + I11 <= 0]
  f4(I12, I13, I14) -> f5(rnd1, I13, I14) [rnd1 = rnd1]
  f3(I15, I16, I17) -> f4(I15, I16, I17) [1 + I17 <= I16]
  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I19 <= I20]
  f2(I21, I22, I23) -> f3(I21, I22, I23) [-1 * I22 + I23 <= 0 /\ -1 * I22 + I23 <= 0]
  f1(I24, I25, I26) -> f2(I24, I25, I26) 

We use the reverse value criterion with the projection function NU:
  NU[f6#(z1,z2,z3)] = z3 + -1 * z2
  NU[f2#(z1,z2,z3)] = z3 + -1 * z2
  NU[f7#(z1,z2,z3)] = z3 + -1 * z2

This gives the following inequalities:
    ==>  I2 + -1 * I1 >= I2 + -1 * I1
  0 <= -1 - I4 + I5  ==>  I5 + -1 * I4 > I5 + -1 * (1 + I4)  with  I5 + -1 * I4 >= 0
    ==>  I8 + -1 * I7 >= I8 + -1 * I7
  I11 <= I10 /\ I10 <= I11 /\ -1 * I10 + I11 <= 0 /\ -1 * I10 + I11 <= 0  ==>  I11 + -1 * I10 > I11 + -1 * (1 + I10)  with  I11 + -1 * I10 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f7#(I0, I1, I2) -> f2#(I0, I1, I2) 
  f6#(I6, I7, I8) -> f2#(I6, I7, I8) 
R = 
  f8(x1, x2, x3) -> f1(x1, x2, x3) 
  f7(I0, I1, I2) -> f2(I0, I1, I2) 
  f2(I3, I4, I5) -> f7(I3, 1 + I4, I5) [0 <= -1 - I4 + I5]
  f6(I6, I7, I8) -> f2(I6, I7, I8) 
  f2(I9, I10, I11) -> f6(I9, 1 + I10, I11) [I11 <= I10 /\ I10 <= I11 /\ -1 * I10 + I11 <= 0 /\ -1 * I10 + I11 <= 0]
  f4(I12, I13, I14) -> f5(rnd1, I13, I14) [rnd1 = rnd1]
  f3(I15, I16, I17) -> f4(I15, I16, I17) [1 + I17 <= I16]
  f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I19 <= I20]
  f2(I21, I22, I23) -> f3(I21, I22, I23) [-1 * I22 + I23 <= 0 /\ -1 * I22 + I23 <= 0]
  f1(I24, I25, I26) -> f2(I24, I25, I26) 

The dependency graph for this problem is:
  1 -> 
  3 -> 
Where:
  1) f7#(I0, I1, I2) -> f2#(I0, I1, I2) 
  3) f6#(I6, I7, I8) -> f2#(I6, I7, I8) 

We have the following SCCs.