/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 =
  init#(x1) -> f1#(rnd1) 
  f3#(I0) -> f3#(I0 + 3) [I0 <= 20]
  f2#(I1) -> f3#(5) [99 <= I1 - 1]
  f2#(I2) -> f2#(I2 + 1) [I2 <= 99]
  f1#(I3) -> f2#(0) 
R = 
  init(x1) -> f1(rnd1) 
  f3(I0) -> f3(I0 + 3) [I0 <= 20]
  f2(I1) -> f3(5) [99 <= I1 - 1]
  f2(I2) -> f2(I2 + 1) [I2 <= 99]
  f1(I3) -> f2(0) 

The dependency graph for this problem is:
  0 -> 4
  1 -> 1
  2 -> 1
  3 -> 2, 3
  4 -> 3
Where:
  0) init#(x1) -> f1#(rnd1) 
  1) f3#(I0) -> f3#(I0 + 3) [I0 <= 20]
  2) f2#(I1) -> f3#(5) [99 <= I1 - 1]
  3) f2#(I2) -> f2#(I2 + 1) [I2 <= 99]
  4) f1#(I3) -> f2#(0) 

We have the following SCCs.
  { 3 }
  { 1 }

DP problem for innermost termination.
P =
  f3#(I0) -> f3#(I0 + 3) [I0 <= 20]
R = 
  init(x1) -> f1(rnd1) 
  f3(I0) -> f3(I0 + 3) [I0 <= 20]
  f2(I1) -> f3(5) [99 <= I1 - 1]
  f2(I2) -> f2(I2 + 1) [I2 <= 99]
  f1(I3) -> f2(0) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1)] = 20 + -1 * z1

This gives the following inequalities:
  I0 <= 20  ==>  20 + -1 * I0 > 20 + -1 * (I0 + 3)  with  20 + -1 * I0 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  f2#(I2) -> f2#(I2 + 1) [I2 <= 99]
R = 
  init(x1) -> f1(rnd1) 
  f3(I0) -> f3(I0 + 3) [I0 <= 20]
  f2(I1) -> f3(5) [99 <= I1 - 1]
  f2(I2) -> f2(I2 + 1) [I2 <= 99]
  f1(I3) -> f2(0) 

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1)] = 99 + -1 * z1

This gives the following inequalities:
  I2 <= 99  ==>  99 + -1 * I2 > 99 + -1 * (I2 + 1)  with  99 + -1 * I2 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.