MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1 - 1, Ar_2)) [ 29 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 >= 30 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 19 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 20 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Ar_2))
		(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_0, Ar_1 - 1, Ar_2)) [ 29 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 >= 30 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 19 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 20 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Ar_2))
		(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(f300) = 1
	Pol(f1) = 0
	Pol(f3) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 20 ]
strictly and produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1 - 1, Ar_2)) [ 29 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_0 >= 30 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 19 >= Ar_1 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 20 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Ar_2))
		(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 + 19 >= 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(f300(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= 20 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 19 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1 - 1, Ar_2)) [ -Ar_1 + 19 >= 0 /\ Ar_0 >= 30 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1 - 1, Ar_2)) [ -Ar_1 + 19 >= 0 /\ 29 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.095 sec (SMT: 1.056 sec)