
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f5(Fresh_2, 0, 0, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9))
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f5(Ar_0, Ar_1 + 1, Ar_2 + 1, 1, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 15 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f5(Ar_0, Ar_1, Ar_2 + 1, 0, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 15 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_1, Ar_1, Fresh_0, Fresh_1, Fresh_1, Fresh_1)) [ Ar_2 >= 16 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_2].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f5(0))
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f5(0))
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 1
	Pol(f0) = 1
	Pol(f5) = 1
	Pol(f27) = 0
orients all transitions weakly and the transition
	f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f5(0))
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 16
	Pol(f0) = 16
	Pol(f5) = -V_1 + 16
	Pol(f27) = -V_1
orients all transitions weakly and the transition
	f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
strictly and produces the following problem:
5:	T:
		(Comp: 1, Cost: 0)     koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)     f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ]
		(Comp: 16, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: 16, Cost: 1)    f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ]
		(Comp: 1, Cost: 1)     f0(Ar_2) -> Com_1(f5(0))
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 34

Time: 0.023 sec (SMT: 0.021 sec)
