MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f2(Fresh_11, Ar_1, Fresh_12, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Fresh_9, Ar_1, Ar_2, Fresh_10, 0, Ar_5, Ar_6)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Fresh_5, Ar_1, Ar_2, Fresh_6, Fresh_7, Fresh_8, Ar_6)) [ 0 >= Fresh_7 + 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Fresh_1, Ar_1, Ar_2, Fresh_2, Fresh_3, Fresh_4, Ar_6)) [ Fresh_3 >= 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 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].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_1, Ar_1)) [ Fresh_3 >= 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_5, Ar_1)) [ 0 >= Fresh_7 + 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_9, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_11, Ar_1)) [ Ar_0 >= Ar_1 ]
	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, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_1, Ar_1)) [ Fresh_3 >= 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_5, Ar_1)) [ 0 >= Fresh_7 + 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_9, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_11, Ar_1)) [ Ar_0 >= Ar_1 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 1
	Pol(f1) = 1
	Pol(f0) = 1
	Pol(f2) = 0
orients all transitions weakly and the transition
	f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_11, Ar_1)) [ Ar_0 >= Ar_1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_1, Ar_1)) [ Fresh_3 >= 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_5, Ar_1)) [ 0 >= Fresh_7 + 1 /\ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Fresh_9, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_11, Ar_1)) [ Ar_0 >= Ar_1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.643 sec (SMT: 0.610 sec)