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

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3, 4
  3 -> 2
  4 -> 5, 6
  5 -> 
  6 -> 
Where:
  0) f7#(x1) -> f6#(x1) 
  1) f6#(I0) -> f4#(0) 
  2) f4#(I2) -> f3#(I2) 
  3) f3#(I3) -> f4#(1 + I3) [1 + I3 <= 2]
  4) f3#(I4) -> f1#(I4) [2 <= I4]
  5) f1#(I5) -> f2#(I5) 
  6) f1#(I6) -> f2#(I6) 

We have the following SCCs.
  { 2, 3 }

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

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

This gives the following inequalities:
    ==>  2 + -1 * (1 + I2) >= 2 + -1 * (1 + I2)
  1 + I3 <= 2  ==>  2 + -1 * (1 + I3) > 2 + -1 * (1 + (1 + I3))  with  2 + -1 * (1 + I3) >= 0

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  2 -> 
Where:
  2) f4#(I2) -> f3#(I2) 

We have the following SCCs.