MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f27(Ar_0, Ar_1) -> Com_1(f29(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 1, Ar_1)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0, Ar_1)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0, 0)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_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

Testing for reachability in the complexity graph removes the following transitions from problem 1:
	f25(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
	f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
	f27(Ar_0, Ar_1) -> Com_1(f29(Ar_0, Ar_1))
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0, Ar_1)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 1, Ar_1)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0, 0)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_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 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0, Ar_1)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 1, Ar_1)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0, 0)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_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 3 to obtain the following invariants:
  For symbol f19: -X_2 >= 0 /\ X_2 >= 0


This yielded 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: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1))
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0, 0)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 1, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 2.637 sec (SMT: 2.554 sec)