/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 =
  f9#(x1) -> f8#(x1) 
  f8#(I0) -> f3#(0) 
  f7#(I1) -> f5#(I1) [2 <= I1]
  f7#(I2) -> f5#(I2) [1 + I2 <= 2]
  f2#(I3) -> f7#(I3) 
  f3#(I5) -> f4#(I5) 
  f4#(I6) -> f2#(I6) 
  f4#(I7) -> f1#(I7) 
  f4#(I8) -> f1#(I8) 
  f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3]
  f1#(I10) -> f2#(I10) [3 <= I10]
R = 
  f9(x1) -> f8(x1) 
  f8(I0) -> f3(0) 
  f7(I1) -> f5(I1) [2 <= I1]
  f7(I2) -> f5(I2) [1 + I2 <= 2]
  f2(I3) -> f7(I3) 
  f5(I4) -> f6(I4) 
  f3(I5) -> f4(I5) 
  f4(I6) -> f2(I6) 
  f4(I7) -> f1(I7) 
  f4(I8) -> f1(I8) 
  f1(I9) -> f3(1 + I9) [1 + I9 <= 3]
  f1(I10) -> f2(I10) [3 <= I10]

The dependency graph for this problem is:
  0 -> 1
  1 -> 5
  2 -> 
  3 -> 
  4 -> 2, 3
  5 -> 6, 7, 8
  6 -> 4
  7 -> 9, 10
  8 -> 9, 10
  9 -> 5
  10 -> 4
Where:
  0) f9#(x1) -> f8#(x1) 
  1) f8#(I0) -> f3#(0) 
  2) f7#(I1) -> f5#(I1) [2 <= I1]
  3) f7#(I2) -> f5#(I2) [1 + I2 <= 2]
  4) f2#(I3) -> f7#(I3) 
  5) f3#(I5) -> f4#(I5) 
  6) f4#(I6) -> f2#(I6) 
  7) f4#(I7) -> f1#(I7) 
  8) f4#(I8) -> f1#(I8) 
  9) f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3]
  10) f1#(I10) -> f2#(I10) [3 <= I10]

We have the following SCCs.
  { 5, 7, 8, 9 }

DP problem for innermost termination.
P =
  f3#(I5) -> f4#(I5) 
  f4#(I7) -> f1#(I7) 
  f4#(I8) -> f1#(I8) 
  f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3]
R = 
  f9(x1) -> f8(x1) 
  f8(I0) -> f3(0) 
  f7(I1) -> f5(I1) [2 <= I1]
  f7(I2) -> f5(I2) [1 + I2 <= 2]
  f2(I3) -> f7(I3) 
  f5(I4) -> f6(I4) 
  f3(I5) -> f4(I5) 
  f4(I6) -> f2(I6) 
  f4(I7) -> f1(I7) 
  f4(I8) -> f1(I8) 
  f1(I9) -> f3(1 + I9) [1 + I9 <= 3]
  f1(I10) -> f2(I10) [3 <= I10]

We use the extended value criterion with the projection function NU:
  NU[f1#(x0)] = -x0 + 2
  NU[f4#(x0)] = -x0 + 2
  NU[f3#(x0)] = -x0 + 2

This gives the following inequalities:
    ==>  -I5 + 2 >= -I5 + 2
    ==>  -I7 + 2 >= -I7 + 2
    ==>  -I8 + 2 >= -I8 + 2
  1 + I9 <= 3  ==>  -I9 + 2 > -(1 + I9) + 2  with  -I9 + 2 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I5) -> f4#(I5) 
  f4#(I7) -> f1#(I7) 
  f4#(I8) -> f1#(I8) 
R = 
  f9(x1) -> f8(x1) 
  f8(I0) -> f3(0) 
  f7(I1) -> f5(I1) [2 <= I1]
  f7(I2) -> f5(I2) [1 + I2 <= 2]
  f2(I3) -> f7(I3) 
  f5(I4) -> f6(I4) 
  f3(I5) -> f4(I5) 
  f4(I6) -> f2(I6) 
  f4(I7) -> f1(I7) 
  f4(I8) -> f1(I8) 
  f1(I9) -> f3(1 + I9) [1 + I9 <= 3]
  f1(I10) -> f2(I10) [3 <= I10]

The dependency graph for this problem is:
  5 -> 7, 8
  7 -> 
  8 -> 
Where:
  5) f3#(I5) -> f4#(I5) 
  7) f4#(I7) -> f1#(I7) 
  8) f4#(I8) -> f1#(I8) 

We have the following SCCs.