WORST_CASE(?, O(1))

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(1, Ar_1, Ar_2, Ar_3)) [ 9 >= E /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(2, Ar_1, Ar_2, 2)) [ 9 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 1:
	f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(1, Ar_1, Ar_2, Ar_3)) [ 9 >= E /\ Ar_0 = 0 ]
	f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(2, Ar_1, Ar_2, 2)) [ 9 >= Ar_0 ]
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f4) = 0
	Pol(f3) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = -V_1 + 10
	Pol(f4) = -V_1 + 10
	Pol(f3) = 10
	Pol(koat_start) = 10
orients all transitions weakly and the transition
	f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]
strictly and produces the following problem:
5:	T:
		(Comp: 1, Cost: 1)     f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]
		(Comp: 10, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]
		(Comp: 1, Cost: 1)     f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))
		(Comp: 1, Cost: 0)     koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 12

Time: 0.449 sec (SMT: 0.436 sec)