MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 >= Ar_0 + Ar_1 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - Ar_1, Ar_1)) [ Ar_0 >= 0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 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)    f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 >= Ar_0 + Ar_1 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - Ar_1, Ar_1)) [ Ar_0 >= 0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f1) = 1
	Pol(f2) = 1
	Pol(f3) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 >= Ar_0 + Ar_1 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - Ar_1, Ar_1)) [ Ar_0 >= 0 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 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 - 2 >= 0 /\ X_1 - X_2 - 3 >= 0 /\ X_1 - 1 >= 0


This yielded 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)    f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ -Ar_1 - 2 >= 0 /\ Ar_0 - Ar_1 - 3 >= 0 /\ Ar_0 - 1 >= 0 /\ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - Ar_1, Ar_1)) [ -Ar_1 - 2 >= 0 /\ Ar_0 - Ar_1 - 3 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 >= Ar_0 + Ar_1 + 1 /\ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for unsatisfiable constraints removes the following transition from problem 4:
	f2(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ -Ar_1 - 2 >= 0 /\ Ar_0 - Ar_1 - 3 >= 0 /\ Ar_0 - 1 >= 0 /\ 0 >= Ar_0 + 1 ]
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - Ar_1, Ar_1)) [ -Ar_1 - 2 >= 0 /\ Ar_0 - Ar_1 - 3 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 >= Ar_0 + Ar_1 + 1 /\ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.028 sec (SMT: 0.979 sec)