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, Ar_10) -> Com_1(f2(Ar_0, Fresh_17, Fresh_18, Fresh_19, Ar_5, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ Ar_0 >= 2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f1(Ar_0, Fresh_14, Fresh_15, Ar_3, Ar_4, Ar_5, Fresh_16, Ar_7, Ar_8, Ar_9, Ar_10)) [ 1 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f2(Ar_0, Fresh_7, Fresh_8, Fresh_9, Fresh_10, Fresh_10, Ar_6, Ar_7, Fresh_11, Fresh_12, Fresh_13)) [ Ar_7 >= 1 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f1(Ar_0, Fresh_0, Fresh_1, Ar_3, Fresh_2, Fresh_3, Fresh_4, Ar_7, Fresh_5, Fresh_2, Fresh_6)) [ 0 >= Ar_7 ]
		(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, Ar_10) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ 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_7].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_7) -> Com_1(f300(Ar_0, Ar_7)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_7) -> Com_1(f1(Ar_0, Ar_7)) [ 0 >= Ar_7 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_7) -> Com_1(f2(Ar_0, Ar_7)) [ Ar_7 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_7) -> Com_1(f1(Ar_0, Ar_7)) [ 1 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_7) -> Com_1(f2(Ar_0, Ar_7)) [ Ar_0 >= 2 ]
	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_7) -> Com_1(f300(Ar_0, Ar_7)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_7) -> Com_1(f1(Ar_0, Ar_7)) [ 0 >= Ar_7 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_7) -> Com_1(f2(Ar_0, Ar_7)) [ Ar_7 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_7) -> Com_1(f1(Ar_0, Ar_7)) [ 1 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_7) -> Com_1(f2(Ar_0, Ar_7)) [ Ar_0 >= 2 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f2: X_2 - 1 >= 0


This yielded the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_7) -> Com_1(f2(Ar_0, Ar_7)) [ Ar_7 - 1 >= 0 /\ Ar_0 >= 2 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_7) -> Com_1(f1(Ar_0, Ar_7)) [ Ar_7 - 1 >= 0 /\ 1 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_7) -> Com_1(f2(Ar_0, Ar_7)) [ Ar_7 >= 1 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_7) -> Com_1(f1(Ar_0, Ar_7)) [ 0 >= Ar_7 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_7) -> Com_1(f300(Ar_0, Ar_7)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.633 sec (SMT: 0.604 sec)