/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 =
  f8#(x1) -> f7#(x1) 
  f7#(I0) -> f2#(0) 
  f2#(I1) -> f5#(I1) 
  f5#(I2) -> f4#(I2) [1 + I2 <= 42]
  f4#(I4) -> f3#(I4) 
  f4#(I5) -> f1#(I5) 
  f4#(I6) -> f3#(I6) 
  f3#(I7) -> f1#(I7) 
  f1#(I8) -> f2#(1 + I8) 
R = 
  f8(x1) -> f7(x1) 
  f7(I0) -> f2(0) 
  f2(I1) -> f5(I1) 
  f5(I2) -> f4(I2) [1 + I2 <= 42]
  f5(I3) -> f6(I3) [42 <= I3]
  f4(I4) -> f3(I4) 
  f4(I5) -> f1(I5) 
  f4(I6) -> f3(I6) 
  f3(I7) -> f1(I7) 
  f1(I8) -> f2(1 + I8) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 4, 5, 6
  4 -> 7
  5 -> 8
  6 -> 7
  7 -> 8
  8 -> 2
Where:
  0) f8#(x1) -> f7#(x1) 
  1) f7#(I0) -> f2#(0) 
  2) f2#(I1) -> f5#(I1) 
  3) f5#(I2) -> f4#(I2) [1 + I2 <= 42]
  4) f4#(I4) -> f3#(I4) 
  5) f4#(I5) -> f1#(I5) 
  6) f4#(I6) -> f3#(I6) 
  7) f3#(I7) -> f1#(I7) 
  8) f1#(I8) -> f2#(1 + I8) 

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7, 8 }

DP problem for innermost termination.
P =
  f2#(I1) -> f5#(I1) 
  f5#(I2) -> f4#(I2) [1 + I2 <= 42]
  f4#(I4) -> f3#(I4) 
  f4#(I5) -> f1#(I5) 
  f4#(I6) -> f3#(I6) 
  f3#(I7) -> f1#(I7) 
  f1#(I8) -> f2#(1 + I8) 
R = 
  f8(x1) -> f7(x1) 
  f7(I0) -> f2(0) 
  f2(I1) -> f5(I1) 
  f5(I2) -> f4(I2) [1 + I2 <= 42]
  f5(I3) -> f6(I3) [42 <= I3]
  f4(I4) -> f3(I4) 
  f4(I5) -> f1(I5) 
  f4(I6) -> f3(I6) 
  f3(I7) -> f1(I7) 
  f1(I8) -> f2(1 + I8) 

We use the extended value criterion with the projection function NU:
  NU[f1#(x0)] = -x0 + 40
  NU[f3#(x0)] = -x0 + 40
  NU[f4#(x0)] = -x0 + 40
  NU[f5#(x0)] = -x0 + 41
  NU[f2#(x0)] = -x0 + 41

This gives the following inequalities:
    ==>  -I1 + 41 >= -I1 + 41
  1 + I2 <= 42  ==>  -I2 + 41 > -I2 + 40  with  -I2 + 41 >= 0
    ==>  -I4 + 40 >= -I4 + 40
    ==>  -I5 + 40 >= -I5 + 40
    ==>  -I6 + 40 >= -I6 + 40
    ==>  -I7 + 40 >= -I7 + 40
    ==>  -I8 + 40 >= -(1 + I8) + 41

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I1) -> f5#(I1) 
  f4#(I4) -> f3#(I4) 
  f4#(I5) -> f1#(I5) 
  f4#(I6) -> f3#(I6) 
  f3#(I7) -> f1#(I7) 
  f1#(I8) -> f2#(1 + I8) 
R = 
  f8(x1) -> f7(x1) 
  f7(I0) -> f2(0) 
  f2(I1) -> f5(I1) 
  f5(I2) -> f4(I2) [1 + I2 <= 42]
  f5(I3) -> f6(I3) [42 <= I3]
  f4(I4) -> f3(I4) 
  f4(I5) -> f1(I5) 
  f4(I6) -> f3(I6) 
  f3(I7) -> f1(I7) 
  f1(I8) -> f2(1 + I8) 

The dependency graph for this problem is:
  2 -> 
  4 -> 7
  5 -> 8
  6 -> 7
  7 -> 8
  8 -> 2
Where:
  2) f2#(I1) -> f5#(I1) 
  4) f4#(I4) -> f3#(I4) 
  5) f4#(I5) -> f1#(I5) 
  6) f4#(I6) -> f3#(I6) 
  7) f3#(I7) -> f1#(I7) 
  8) f1#(I8) -> f2#(1 + I8) 

We have the following SCCs.