
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

By chaining the transition koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] with all transitions in problem 4, the following new transition is obtained:
	koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_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

Testing for reachability in the complexity graph removes the following transition from problem 5:
	f0(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1))
We thus obtain the following problem:
6:	T:
		(Comp: ?, Cost: 1)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 ] with all transitions in problem 6, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 ]
We thus obtain the following problem:
7:	T:
		(Comp: ?, Cost: 2)    f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 ] with all transitions in problem 7, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 ]
We thus obtain the following problem:
8:	T:
		(Comp: ?, Cost: 3)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 ] with all transitions in problem 8, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 1, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 ]
We thus obtain the following problem:
9:	T:
		(Comp: ?, Cost: 4)    f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 1, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 1, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 ] with all transitions in problem 9, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 2, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 ]
We thus obtain the following problem:
10:	T:
		(Comp: ?, Cost: 5)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 2, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 2, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 ] with all transitions in problem 10, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 2, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 ]
We thus obtain the following problem:
11:	T:
		(Comp: ?, Cost: 6)    f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 2, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 2, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 ] with all transitions in problem 11, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 3, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 ]
We thus obtain the following problem:
12:	T:
		(Comp: ?, Cost: 7)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 3, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 3, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 ] with all transitions in problem 12, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 3, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
We thus obtain the following problem:
13:	T:
		(Comp: ?, Cost: 8)    f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 3, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f19(Ar_0, Ar_1) -> Com_1(f10(Ar_0 + 3, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ] with all transitions in problem 13, the following new transition is obtained:
	f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
We thus obtain the following problem:
14:	T:
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0, 0)) [ Ar_0 >= 6 ] with all transitions in problem 14, the following new transition is obtained:
	f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 1, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 ]
We thus obtain the following problem:
15:	T:
		(Comp: ?, Cost: 2)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 1, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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(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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 1, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 ] with all transitions in problem 15, the following new transition is obtained:
	f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 2, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 ]
We thus obtain the following problem:
16:	T:
		(Comp: ?, Cost: 3)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 2, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 ]
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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(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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 2, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 ] with all transitions in problem 16, the following new transition is obtained:
	f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 3, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 ]
We thus obtain the following problem:
17:	T:
		(Comp: ?, Cost: 4)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 3, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 ]
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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(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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 3, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 ] with all transitions in problem 17, the following new transition is obtained:
	f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 4, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 /\ Ar_0 - 3 >= 3 ]
We thus obtain the following problem:
18:	T:
		(Comp: ?, Cost: 5)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 4, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 /\ Ar_0 - 3 >= 3 ]
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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(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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ] with all transitions in problem 18, the following new transition is obtained:
	f9(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
We thus obtain the following problem:
19:	T:
		(Comp: ?, Cost: 2)    f9(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 5)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 4, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 /\ Ar_0 - 3 >= 3 ]
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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(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)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ] with all transitions in problem 19, the following new transition is obtained:
	f9(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ]
We thus obtain the following problem:
20:	T:
		(Comp: ?, Cost: 2)    f9(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ]
		(Comp: ?, Cost: 2)    f9(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 5)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 4, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 /\ Ar_0 - 3 >= 3 ]
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    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(f10(Ar_0, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 20:
	f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ Ar_0 >= 6 ]
We thus obtain the following problem:
21:	T:
		(Comp: ?, Cost: 9)    f19(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 4, Ar_1)) [ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 2 >= Ar_0 /\ 5 >= Ar_0 /\ 5 >= Ar_0 + 1 /\ 5 >= Ar_0 + 2 /\ 5 >= Ar_0 + 3 ]
		(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)    f10(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 2)    f9(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Fresh_2)) [ Ar_0 >= 6 /\ 0 >= Fresh_2 + 1 ]
		(Comp: ?, Cost: 2)    f9(Ar_0, Ar_1) -> Com_1(f9(Ar_0 + 1, Fresh_1)) [ Ar_0 >= 6 /\ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 5)    f9(Ar_0, Ar_1) -> Com_1(f19(Ar_0 - 4, 0)) [ Ar_0 >= 6 /\ 0 >= 0 /\ Ar_0 >= 3 /\ Ar_0 - 1 >= 3 /\ Ar_0 - 2 >= 3 /\ Ar_0 - 3 >= 3 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1) -> Com_1(f10(Ar_0, Ar_1)) [ 5 >= Ar_0 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f9(Fresh_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.390 sec (SMT: 0.331 sec)
