/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 =
  f5#(x1, x2) -> f4#(x1, x2) 
  f4#(I0, I1) -> f3#(I0, 0) 
  f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
  f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, 0) 
  f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
  f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20]
  f1(I6, I7) -> f2(I6, I7) [20 <= I7]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2) -> f4#(x1, x2) 
  1) f4#(I0, I1) -> f3#(I0, 0) 
  2) f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
  3) f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
  f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, 0) 
  f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
  f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20]
  f1(I6, I7) -> f2(I6, I7) [20 <= I7]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2)] = 20 + -1 * (1 + z2)
  NU[f3#(z1,z2)] = 20 + -1 * (1 + (2 + z2))

This gives the following inequalities:
  rnd1 = 2 + 2 + I3  ==>  20 + -1 * (1 + (2 + I3)) >= 20 + -1 * (1 + (2 + I3))
  1 + I5 <= 20  ==>  20 + -1 * (1 + I5) > 20 + -1 * (1 + (2 + I5))  with  20 + -1 * (1 + I5) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f3(I0, 0) 
  f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
  f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20]
  f1(I6, I7) -> f2(I6, I7) [20 <= I7]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]

We have the following SCCs.