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

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 4, 5
  3 -> 4, 5
  4 -> 3
  5 -> 2
Where:
  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  1) f4#(I0, I1, I2) -> f2#(0, I1, I2) 
  2) f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5]
  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
  4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
  5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13]

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

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

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

This gives the following inequalities:
  1 + I3 <= I5  ==>  -I3 + I5 - 1 > -I3 + I5 - 2  with  -I3 + I5 - 1 >= 0
    ==>  -I6 + I8 - 2 >= -I6 + I8 - 2
  1 + I10 <= I9  ==>  -I9 + I11 - 2 >= -I9 + I11 - 2
  I12 <= I13  ==>  -I12 + I14 - 2 >= -(1 + I12) + I14 - 1

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  3 -> 4, 5
  4 -> 3
  5 -> 
Where:
  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
  4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9]
  5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13]

We have the following SCCs.
  { 3, 4 }

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

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

This gives the following inequalities:
    ==>  I6 + -1 * (1 + I7) >= I6 + -1 * (1 + I7)
  1 + I10 <= I9  ==>  I9 + -1 * (1 + I10) > I9 + -1 * (1 + (1 + I10))  with  I9 + -1 * (1 + I10) >= 0

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  3 -> 
Where:
  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 

We have the following SCCs.