/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/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) -> f4#(0, I1, I2) 
  f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0]
  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f4(0, I1, I2) 
  f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4]
  f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0]
  f3(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14]
  f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 4
  4 -> 3
Where:
  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  1) f6#(I0, I1, I2) -> f4#(0, I1, I2) 
  2) f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0]
  3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
  4) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14]

We have the following SCCs.
  { 3, 4 }

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

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = z3
  NU[f3#(z1,z2,z3)] = z3

This gives the following inequalities:
    ==>  I11 (>! \union =) I11
  1 <= I14  ==>  I14 >! -1 + I14

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  3 -> 
Where:
  3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 

We have the following SCCs.