MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f300(Ar_0, Fresh_6, Fresh_7, Fresh_8, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f2(Ar_0, Fresh_3, Fresh_4, Ar_3, Fresh_5, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_0, Fresh_1, Fresh_2, Fresh_2, Fresh_0))
		(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(f1(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_0].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f1(Ar_0)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0) -> Com_1(f2(Ar_0))
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f300(Ar_0)) [ 0 >= Ar_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_0) -> Com_1(f1(Ar_0)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0) -> Com_1(f2(Ar_0))
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0) -> Com_1(f300(Ar_0)) [ 0 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.351 sec (SMT: 0.329 sec)