/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 =
  f7#(x1) -> f6#(x1) 
  f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1]
  f4#(I1) -> f3#(I1) 
  f5#(I2) -> f4#(I2) [1 <= I2]
  f5#(I3) -> f1#(I3) [I3 <= 0]
  f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20]
  f3#(I5) -> f1#(I5) [20 <= I5]
R = 
  f7(x1) -> f6(x1) 
  f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1]
  f4(I1) -> f3(I1) 
  f5(I2) -> f4(I2) [1 <= I2]
  f5(I3) -> f1(I3) [I3 <= 0]
  f3(I4) -> f4(1 + I4) [1 + I4 <= 20]
  f3(I5) -> f1(I5) [20 <= I5]
  f1(I6) -> f2(I6) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 3, 4
  2 -> 5, 6
  3 -> 2
  4 -> 
  5 -> 2
  6 -> 
Where:
  0) f7#(x1) -> f6#(x1) 
  1) f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1]
  2) f4#(I1) -> f3#(I1) 
  3) f5#(I2) -> f4#(I2) [1 <= I2]
  4) f5#(I3) -> f1#(I3) [I3 <= 0]
  5) f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20]
  6) f3#(I5) -> f1#(I5) [20 <= I5]

We have the following SCCs.
  { 2, 5 }

DP problem for innermost termination.
P =
  f4#(I1) -> f3#(I1) 
  f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20]
R = 
  f7(x1) -> f6(x1) 
  f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1]
  f4(I1) -> f3(I1) 
  f5(I2) -> f4(I2) [1 <= I2]
  f5(I3) -> f1(I3) [I3 <= 0]
  f3(I4) -> f4(1 + I4) [1 + I4 <= 20]
  f3(I5) -> f1(I5) [20 <= I5]
  f1(I6) -> f2(I6) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1)] = 20 + -1 * (1 + z1)
  NU[f4#(z1)] = 20 + -1 * (1 + z1)

This gives the following inequalities:
    ==>  20 + -1 * (1 + I1) >= 20 + -1 * (1 + I1)
  1 + I4 <= 20  ==>  20 + -1 * (1 + I4) > 20 + -1 * (1 + (1 + I4))  with  20 + -1 * (1 + I4) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I1) -> f3#(I1) 
R = 
  f7(x1) -> f6(x1) 
  f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1]
  f4(I1) -> f3(I1) 
  f5(I2) -> f4(I2) [1 <= I2]
  f5(I3) -> f1(I3) [I3 <= 0]
  f3(I4) -> f4(1 + I4) [1 + I4 <= 20]
  f3(I5) -> f1(I5) [20 <= I5]
  f1(I6) -> f2(I6) 

The dependency graph for this problem is:
  2 -> 
Where:
  2) f4#(I1) -> f3#(I1) 

We have the following SCCs.