/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 =
  f6#(x1) -> f5#(x1) 
  f5#(I0) -> f2#(rnd1) [y1 = 5 /\ rnd1 = rnd1]
  f2#(I1) -> f3#(I1) 
  f3#(I2) -> f1#(I2) [1 + I2 <= 4]
  f1#(I4) -> f2#(1 + I4) [1 <= I4]
  f1#(I5) -> f2#(1) [I5 <= 0]
R = 
  f6(x1) -> f5(x1) 
  f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1]
  f2(I1) -> f3(I1) 
  f3(I2) -> f1(I2) [1 + I2 <= 4]
  f3(I3) -> f4(I3) [4 <= I3]
  f1(I4) -> f2(1 + I4) [1 <= I4]
  f1(I5) -> f2(1) [I5 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 4, 5
  4 -> 2
  5 -> 2
Where:
  0) f6#(x1) -> f5#(x1) 
  1) f5#(I0) -> f2#(rnd1) [y1 = 5 /\ rnd1 = rnd1]
  2) f2#(I1) -> f3#(I1) 
  3) f3#(I2) -> f1#(I2) [1 + I2 <= 4]
  4) f1#(I4) -> f2#(1 + I4) [1 <= I4]
  5) f1#(I5) -> f2#(1) [I5 <= 0]

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

DP problem for innermost termination.
P =
  f2#(I1) -> f3#(I1) 
  f3#(I2) -> f1#(I2) [1 + I2 <= 4]
  f1#(I4) -> f2#(1 + I4) [1 <= I4]
  f1#(I5) -> f2#(1) [I5 <= 0]
R = 
  f6(x1) -> f5(x1) 
  f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1]
  f2(I1) -> f3(I1) 
  f3(I2) -> f1(I2) [1 + I2 <= 4]
  f3(I3) -> f4(I3) [4 <= I3]
  f1(I4) -> f2(1 + I4) [1 <= I4]
  f1(I5) -> f2(1) [I5 <= 0]

We use the extended value criterion with the projection function NU:
  NU[f1#(x0)] = -x0 + 2
  NU[f3#(x0)] = -x0 + 3
  NU[f2#(x0)] = -x0 + 3

This gives the following inequalities:
    ==>  -I1 + 3 >= -I1 + 3
  1 + I2 <= 4  ==>  -I2 + 3 > -I2 + 2  with  -I2 + 3 >= 0
  1 <= I4  ==>  -I4 + 2 >= -(1 + I4) + 3
  I5 <= 0  ==>  -I5 + 2 >= -1 + 3

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I1) -> f3#(I1) 
  f1#(I4) -> f2#(1 + I4) [1 <= I4]
  f1#(I5) -> f2#(1) [I5 <= 0]
R = 
  f6(x1) -> f5(x1) 
  f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1]
  f2(I1) -> f3(I1) 
  f3(I2) -> f1(I2) [1 + I2 <= 4]
  f3(I3) -> f4(I3) [4 <= I3]
  f1(I4) -> f2(1 + I4) [1 <= I4]
  f1(I5) -> f2(1) [I5 <= 0]

The dependency graph for this problem is:
  2 -> 
  4 -> 2
  5 -> 2
Where:
  2) f2#(I1) -> f3#(I1) 
  4) f1#(I4) -> f2#(1 + I4) [1 <= I4]
  5) f1#(I5) -> f2#(1) [I5 <= 0]

We have the following SCCs.