MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f1(0, 0))
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1))
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f30(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f20(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)    f20(Ar_0, Ar_1) -> Com_1(f1(0, 0))
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1))
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f30(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f20(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f20) = 1
	Pol(f1) = 1
	Pol(f30) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f1(Ar_0, Ar_1) -> Com_1(f30(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f1(0, 0))
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1))
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f30(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f20(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 f1: X_1 - X_2 >= 0 /\ X_2 >= 0 /\ X_1 + X_2 >= 0 /\ -X_1 + X_2 >= 0 /\ X_1 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f20(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f30(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1)) [ Ar_0 - Ar_1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ]
		(Comp: 1, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f1(0, 0))
	start location:	koat_start
	leaf cost:	0

Testing for unsatisfiable constraints removes the following transition from problem 4:
	f1(Ar_0, Ar_1) -> Com_1(f30(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_0 >= Ar_1 + 1 ]
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f20(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1)) [ Ar_0 - Ar_1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ]
		(Comp: 1, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f1(0, 0))
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.729 sec (SMT: 0.698 sec)