MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_2, Ar_2 - 1, Ar_0, Ar_4, Ar_5)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3, Ar_2, Ar_0)) [ Ar_0 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Fresh_0, Ar_3, Ar_4, Ar_5)) [ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 1:
	f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3, Ar_2, Ar_0)) [ Ar_0 >= 1 /\ Ar_2 >= 1 ]
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_2, Ar_2 - 1, Ar_0, Ar_4, Ar_5)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Fresh_0, Ar_3, Ar_4, Ar_5)) [ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_2, Ar_2 - 1, Ar_0, Ar_4, Ar_5)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Fresh_0, Ar_3, Ar_4, Ar_5)) [ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f0: -X_1 + X_3 + 4999 >= 0 /\ -X_1 + 5000 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Fresh_0, Ar_3, Ar_4, Ar_5)) [ Fresh_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_2, Ar_2 - 1, Ar_0, Ar_4, Ar_5)) [ -Ar_0 + Ar_2 + 4999 >= 0 /\ -Ar_0 + 5000 >= 0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(5000, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ -Ar_0 + Ar_2 + 4999 >= 0 /\ -Ar_0 + 5000 >= 0 /\ 0 >= Ar_0 /\ Ar_2 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.399 sec (SMT: 1.347 sec)