/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 =
  f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
  f5#(I0, I1, I2) -> f1#(I0, I1, I2) 
  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f1(I0, I1, I2) 
  f4(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3, 5
  2 -> 3, 5
  3 -> 2
  4 -> 3, 5
  5 -> 4
Where:
  0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
  1) f5#(I0, I1, I2) -> f1#(I0, I1, I2) 
  2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
  5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]

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

DP problem for innermost termination.
P =
  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f1(I0, I1, I2) 
  f4(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2,z3)] = 2 - z2 + -1 * 0
  NU[f1#(z1,z2,z3)] = 2 - z2 + -1 * 0
  NU[f4#(z1,z2,z3)] = 2 - z2 + -1 * 0

This gives the following inequalities:
    ==>  2 - I4 + -1 * 0 >= 2 - I4 + -1 * 0
  0 <= 1 - I8  ==>  2 - I7 + -1 * 0 >= 2 - (1 + I7) + -1 * 0
    ==>  2 - I10 + -1 * 0 >= 2 - I10 + -1 * 0
  0 <= 2 - I13 /\ 2 - I14 <= 0  ==>  2 - I13 + -1 * 0 > 2 - (1 + I13) + -1 * 0  with  2 - I13 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f1(I0, I1, I2) 
  f4(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0]

The dependency graph for this problem is:
  2 -> 3
  3 -> 2
  4 -> 3
Where:
  2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f1(I0, I1, I2) 
  f4(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = 1 - z3 + -1 * 0
  NU[f4#(z1,z2,z3)] = 1 - z3 + -1 * 0

This gives the following inequalities:
    ==>  1 - I5 + -1 * 0 >= 1 - I5 + -1 * 0
  0 <= 1 - I8  ==>  1 - I8 + -1 * 0 > 1 - (1 + I8) + -1 * 0  with  1 - I8 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f1(I0, I1, I2) 
  f4(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8]
  f3(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 

We have the following SCCs.