
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(400, 0, 0, Ar_3, Ar_4, Ar_5))
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(Ar_0, Ar_1, 0, 0, 0, Ar_5)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_4, Fresh_5, Ar_4, Fresh_4))
		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_2, Fresh_3, Ar_4, Fresh_2))
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0, Fresh_1, Ar_4, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_0, Ar_1, Ar_2].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_2))
		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_4))
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(400, 0, 0))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_2))
	f7(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_4))
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(400, 0, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 3 produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ]
		(Comp: 1, Cost: 1)    f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(400, 0, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f5) = 1
	Pol(f11) = 0
	Pol(f10) = 0
	Pol(f12) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transitions
	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ]
	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]
	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ]
strictly and produces the following problem:
5:	T:
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ]
		(Comp: 1, Cost: 1)    f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(400, 0, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f5) = V_1 - V_2
	Pol(f11) = V_1 - V_2
	Pol(f10) = V_1 - V_2
	Pol(f12) = 400
	Pol(koat_start) = 400
orients all transitions weakly and the transition
	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ]
strictly and produces the following problem:
6:	T:
		(Comp: 1, Cost: 1)      f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ]
		(Comp: 1, Cost: 1)      f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)      f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: 400, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ]
		(Comp: 1, Cost: 1)      f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(400, 0, 0))
		(Comp: 1, Cost: 0)      koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 404

Time: 0.021 sec (SMT: 0.017 sec)
