MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 - Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 1 produces the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 - Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f4) = 1
	Pol(f6) = 0
	Pol(f5) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]
strictly and produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 - Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 0 ]
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f4: -X_2 - 1 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ -Ar_1 - 1 >= 0 /\ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 - Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ -Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.782 sec (SMT: 1.727 sec)