WORST_CASE(?, O(n^1))

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ Ar_2 >= 2 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ Ar_2 >= 2 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 1:
	f0(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0, Ar_1, Ar_2))
	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ Ar_2 >= 2 ]
	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ 0 >= Ar_2 ]
	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 /\ Ar_2 >= 2 ]
	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]
	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ 0 >= Ar_0 + 1 /\ Ar_2 = 1 ]
	f5(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f4) = 1
	Pol(f7) = 0
	Pol(f3) = 1
	Pol(f6) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f4) = 1
	Pol(f7) = 0
	Pol(f3) = 1
	Pol(f6) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
strictly and produces the following problem:
5:	T:
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f4) = V_1
	Pol(f7) = 0
	Pol(f3) = -V_1
	Pol(f6) = V_1
	Pol(koat_start) = V_1
orients all transitions weakly and the transition
	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
strictly and produces the following problem:
6:	T:
		(Comp: 1, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)       f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: Ar_0, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 6 produces the following problem:
7:	T:
		(Comp: 1, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_0, Ar_2)) [ Ar_0 = 0 ]
		(Comp: Ar_0, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(-Ar_0 - 1, Ar_1, 1)) [ 0 >= Ar_0 + 1 /\ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1, Ar_2) -> Com_1(f7(0, Fresh_1, Ar_2)) [ Ar_0 = 0 ]
		(Comp: Ar_0, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f3(-Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 1 /\ Ar_2 = 1 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 2*Ar_0 + 3

Time: 1.060 sec (SMT: 1.033 sec)