MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_4, Ar_2)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_3, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f17(0, Ar_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, 0)) [ Ar_2 = 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f17(Ar_0, 0, 1)) [ Ar_1 = 0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, Ar_1, Ar_2)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_1, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_0, Ar_1, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 1 produces the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_4, Ar_2)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_3, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f17(0, Ar_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, 0)) [ Ar_2 = 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f17(Ar_0, 0, 1)) [ Ar_1 = 0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, Ar_1, Ar_2)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_1, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_0, Ar_1, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f6) = 2
	Pol(f9) = 2
	Pol(f17) = 1
	Pol(f24) = 0
	Pol(f0) = 2
	Pol(koat_start) = 2
orients all transitions weakly and the transition
	f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, 0)) [ Ar_2 = 1 ]
strictly and produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_4, Ar_2)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_3, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f17(0, Ar_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, 0)) [ Ar_2 = 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f17(Ar_0, 0, 1)) [ Ar_1 = 0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, Ar_1, Ar_2)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_1, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_0, Ar_1, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f6) = 2
	Pol(f9) = 2
	Pol(f17) = 1
	Pol(f24) = 0
	Pol(f0) = 2
	Pol(koat_start) = 2
orients all transitions weakly and the transitions
	f9(Ar_0, Ar_1, Ar_2) -> Com_1(f17(Ar_0, 0, 1)) [ Ar_1 = 0 ]
	f6(Ar_0, Ar_1, Ar_2) -> Com_1(f17(0, Ar_1, Ar_2)) [ Ar_0 = 0 ]
	f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ]
	f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]
strictly and produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_4, Ar_2)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_3, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: 2, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f17(0, Ar_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, 0)) [ Ar_2 = 1 ]
		(Comp: 2, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f17(Ar_0, 0, 1)) [ Ar_1 = 0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, Ar_1, Ar_2)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_1, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_0, Ar_1, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f17: -X_3 + 1 >= 0 /\ X_3 >= 0
  For symbol f6: -X_3 >= 0 /\ X_3 >= 0
  For symbol f9: -X_3 >= 0 /\ X_3 >= 0


This yielded the following problem:
5:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_0, Ar_1, 0))
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_1, Ar_1, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, Ar_1, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ 0 >= Ar_1 + 1 ]
		(Comp: 2, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f17(Ar_0, 0, 1)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_1 = 0 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, 0)) [ -Ar_2 + 1 >= 0 /\ Ar_2 >= 0 /\ Ar_2 = 1 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ -Ar_2 + 1 >= 0 /\ Ar_2 >= 0 /\ Ar_2 >= 2 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ -Ar_2 + 1 >= 0 /\ Ar_2 >= 0 /\ 0 >= Ar_2 ]
		(Comp: 2, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f17(0, Ar_1, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_3, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_4, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ 0 >= Ar_0 + 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for unsatisfiable constraints removes the following transition from problem 5:
	f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ -Ar_2 + 1 >= 0 /\ Ar_2 >= 0 /\ Ar_2 >= 2 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_0, Ar_1, 0))
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_1, Ar_1, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, Ar_1, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ 0 >= Ar_1 + 1 ]
		(Comp: 2, Cost: 1)    f9(Ar_0, Ar_1, Ar_2) -> Com_1(f17(Ar_0, 0, 1)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_1 = 0 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, 0)) [ -Ar_2 + 1 >= 0 /\ Ar_2 >= 0 /\ Ar_2 = 1 ]
		(Comp: 2, Cost: 1)    f17(Ar_0, Ar_1, Ar_2) -> Com_1(f24(Ar_0, Ar_1, Ar_2)) [ -Ar_2 + 1 >= 0 /\ Ar_2 >= 0 /\ 0 >= Ar_2 ]
		(Comp: 2, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f17(0, Ar_1, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_3, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f9(Ar_0, Fresh_4, Ar_2)) [ -Ar_2 >= 0 /\ Ar_2 >= 0 /\ 0 >= Ar_0 + 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 2.345 sec (SMT: 2.273 sec)