WORST_CASE(?, O(n^1))

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, -Ar_0 + Ar_1)) [ Ar_1 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(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)    f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, -Ar_0 + Ar_1)) [ Ar_1 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 2 to obtain the following invariants:
  For symbol f1: X_1 - 1 >= 0


This yielded the following problem:
3:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, -Ar_0 + Ar_1)) [ Ar_0 - 1 >= 0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = V_2
	Pol(f0) = V_2
	Pol(f1) = V_2
orients all transitions weakly and the transition
	f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, -Ar_0 + Ar_1)) [ Ar_0 - 1 >= 0 /\ Ar_1 >= 1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: Ar_1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, -Ar_0 + Ar_1)) [ Ar_0 - 1 >= 0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound Ar_1 + 1

Time: 0.961 sec (SMT: 0.937 sec)