MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 - 1, Ar_1, Ar_2 - 1)) [ Ar_0 >= 1 /\ Ar_1 + 1 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1 + Ar_2 - 1, Ar_2 - 1)) [ Ar_0 >= 0 /\ Ar_1 >= 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 1 produces the following problem:
2:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 - 1, Ar_1, Ar_2 - 1)) [ Ar_0 >= 1 /\ Ar_1 + 1 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1 + Ar_2 - 1, Ar_2 - 1)) [ Ar_0 >= 0 /\ Ar_1 >= 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = V_1
	Pol(f2) = V_1
	Pol(koat_start) = V_1
orients all transitions weakly and the transition
	f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 - 1, Ar_1, Ar_2 - 1)) [ Ar_0 >= 1 /\ Ar_1 + 1 >= Ar_2 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_2 ]
		(Comp: Ar_0, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 - 1, Ar_1, Ar_2 - 1)) [ Ar_0 >= 1 /\ Ar_1 + 1 >= Ar_2 ]
		(Comp: ?, Cost: 1)       f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1 + Ar_2 - 1, Ar_2 - 1)) [ Ar_0 >= 0 /\ Ar_1 >= 0 ]
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f2: X_2 - X_3 >= 0 /\ X_1 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)       f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1 + Ar_2 - 1, Ar_2 - 1)) [ Ar_1 - Ar_2 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 0 ]
		(Comp: Ar_0, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 - 1, Ar_1, Ar_2 - 1)) [ Ar_1 - Ar_2 >= 0 /\ Ar_0 >= 0 /\ Ar_0 >= 1 /\ Ar_1 + 1 >= Ar_2 ]
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_2 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.855 sec (SMT: 1.800 sec)