/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 =
  f5#(x1, x2) -> f4#(x1, x2) 
  f4#(I0, I1) -> f3#(I0, I1) 
  f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
  f2#(I6, I7) -> f1#(I6, I7) 
  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, I1) 
  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
  f2(I6, I7) -> f1(I6, I7) 
  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3, 5
  3 -> 2
  4 -> 3, 5
  5 -> 4
Where:
  0) f5#(x1, x2) -> f4#(x1, x2) 
  1) f4#(I0, I1) -> f3#(I0, I1) 
  2) f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
  4) f2#(I6, I7) -> f1#(I6, I7) 
  5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]

We have the following SCCs.
  { 2, 3, 4, 5 }

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
  f2#(I6, I7) -> f1#(I6, I7) 
  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, I1) 
  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
  f2(I6, I7) -> f1(I6, I7) 
  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z1
  NU[f1#(z1,z2)] = z1
  NU[f3#(z1,z2)] = z1

This gives the following inequalities:
  rnd2 = rnd2 /\ 1 <= I2  ==>  I2 >! -1 + I2
  I5 <= 0  ==>  I4 (>! \union =) I4
    ==>  I6 (>! \union =) I6
  1 <= I9  ==>  I8 (>! \union =) I8

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
  f2#(I6, I7) -> f1#(I6, I7) 
  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, I1) 
  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
  f2(I6, I7) -> f1(I6, I7) 
  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]

The dependency graph for this problem is:
  3 -> 
  4 -> 3, 5
  5 -> 4
Where:
  3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0]
  4) f2#(I6, I7) -> f1#(I6, I7) 
  5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]

We have the following SCCs.
  { 4, 5 }

DP problem for innermost termination.
P =
  f2#(I6, I7) -> f1#(I6, I7) 
  f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, I1) 
  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
  f2(I6, I7) -> f1(I6, I7) 
  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2)] = z2
  NU[f2#(z1,z2)] = z2

This gives the following inequalities:
    ==>  I7 (>! \union =) I7
  1 <= I9  ==>  I9 >! -1 + I9

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I6, I7) -> f1#(I6, I7) 
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, I1) 
  f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2]
  f1(I4, I5) -> f3(I4, I5) [I5 <= 0]
  f2(I6, I7) -> f1(I6, I7) 
  f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9]

The dependency graph for this problem is:
  4 -> 
Where:
  4) f2#(I6, I7) -> f1#(I6, I7) 

We have the following SCCs.