
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(-Ar_1 + Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_0)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f3(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: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(-Ar_1 + Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_1)) [ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_0)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f3(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f2) = 1
	Pol(f3) = 1
	Pol(f4) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f2(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_1)) [ 0 >= Ar_0 ]
strictly and produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(-Ar_1 + Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_1)) [ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_0)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f3(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 - 1 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f3(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_0)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_1)) [ Ar_1 - 1 >= 0 /\ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(-Ar_1 + Ar_0, Ar_1 + 1, Ar_2)) [ Ar_1 - 1 >= 0 /\ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = V_1
	Pol(f3) = V_1
	Pol(f4) = V_1
	Pol(f2) = V_1
orients all transitions weakly and the transition
	f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(-Ar_1 + Ar_0, Ar_1 + 1, Ar_2)) [ Ar_1 - 1 >= 0 /\ Ar_0 >= 1 ]
strictly and produces the following problem:
5:	T:
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f3(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_0)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 1)       f2(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Fresh_1)) [ Ar_1 - 1 >= 0 /\ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 1 ]
		(Comp: Ar_0, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(-Ar_1 + Ar_0, Ar_1 + 1, Ar_2)) [ Ar_1 - 1 >= 0 /\ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound Ar_0 + 3

Time: 0.062 sec (SMT: 0.051 sec)
