MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Fresh_8, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2 - 1, Fresh_6, 0, Fresh_7)) [ Ar_0 >= 1 /\ Ar_2 >= 3 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Fresh_3, Fresh_4, Fresh_5, Ar_5)) [ 0 >= Fresh_5 + 1 /\ Ar_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_1, Fresh_0, Fresh_1, Fresh_2, Ar_5)) [ Fresh_2 >= 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f1(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 1 produces the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Fresh_8, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2 - 1, Fresh_6, 0, Fresh_7)) [ Ar_0 >= 1 /\ Ar_2 >= 3 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Fresh_3, Fresh_4, Fresh_5, Ar_5)) [ 0 >= Fresh_5 + 1 /\ Ar_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_1, Fresh_0, Fresh_1, Fresh_2, Ar_5)) [ Fresh_2 >= 1 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f2) = 0
	Pol(f1) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Fresh_8, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Fresh_8, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2 - 1, Fresh_6, 0, Fresh_7)) [ Ar_0 >= 1 /\ Ar_2 >= 3 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Fresh_3, Fresh_4, Fresh_5, Ar_5)) [ 0 >= Fresh_5 + 1 /\ Ar_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_1, Fresh_0, Fresh_1, Fresh_2, Ar_5)) [ Fresh_2 >= 1 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = V_1
	Pol(f2) = V_1
	Pol(f1) = V_1
	Pol(koat_start) = V_1
orients all transitions weakly and the transitions
	f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Fresh_3, Fresh_4, Fresh_5, Ar_5)) [ 0 >= Fresh_5 + 1 /\ Ar_0 >= 1 ]
	f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Fresh_0, Fresh_1, Fresh_2, Ar_5)) [ Fresh_2 >= 1 /\ Ar_0 >= 1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Fresh_8, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)       f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2 - 1, Fresh_6, 0, Fresh_7)) [ Ar_0 >= 1 /\ Ar_2 >= 3 ]
		(Comp: Ar_0, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Fresh_3, Fresh_4, Fresh_5, Ar_5)) [ 0 >= Fresh_5 + 1 /\ Ar_0 >= 1 ]
		(Comp: Ar_0, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 - 1, Ar_1, Fresh_0, Fresh_1, Fresh_2, Ar_5)) [ Fresh_2 >= 1 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)       f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.644 sec (SMT: 1.574 sec)