/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 =
  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  f6#(I0, I1, I2) -> f3#(0, I1, rnd3) [rnd3 = rnd3]
  f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 10]
  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
  f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 10]
  f1#(I18, I19, I20) -> f2#(I18, 0, I20) [10 <= I18]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
  f2(I9, I10, I11) -> f4(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]

The dependency graph for this problem is:
  0 -> 1
  1 -> 4
  2 -> 3
  3 -> 2
  4 -> 5, 6
  5 -> 4
  6 -> 3
Where:
  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  1) f6#(I0, I1, I2) -> f3#(0, I1, rnd3) [rnd3 = rnd3]
  2) f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 10]
  3) f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
  5) f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 10]
  6) f1#(I18, I19, I20) -> f2#(I18, 0, I20) [10 <= I18]

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

DP problem for innermost termination.
P =
  f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 10]
  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
  f2(I9, I10, I11) -> f4(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = 10 + -1 * (1 + z2)
  NU[f4#(z1,z2,z3)] = 10 + -1 * (1 + z2)

This gives the following inequalities:
  1 + I4 <= 10  ==>  10 + -1 * (1 + I4) > 10 + -1 * (1 + (1 + I4))  with  10 + -1 * (1 + I4) >= 0
    ==>  10 + -1 * (1 + I10) >= 10 + -1 * (1 + I10)

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
  f2(I9, I10, I11) -> f4(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]

The dependency graph for this problem is:
  3 -> 
Where:
  3) f2#(I9, I10, I11) -> f4#(I9, I10, I11) 

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
  f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 10]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
  f2(I9, I10, I11) -> f4(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = 10 + -1 * (1 + z1)
  NU[f3#(z1,z2,z3)] = 10 + -1 * (1 + z1)

This gives the following inequalities:
    ==>  10 + -1 * (1 + I12) >= 10 + -1 * (1 + I12)
  1 + I15 <= 10  ==>  10 + -1 * (1 + I15) > 10 + -1 * (1 + (1 + I15))  with  10 + -1 * (1 + I15) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(0, I1, rnd3) [rnd3 = rnd3]
  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 10]
  f4(I6, I7, I8) -> f5(I6, I7, I8) [10 <= I7]
  f2(I9, I10, I11) -> f4(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 10]
  f1(I18, I19, I20) -> f2(I18, 0, I20) [10 <= I18]

The dependency graph for this problem is:
  4 -> 
Where:
  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 

We have the following SCCs.