
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f26(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    f27(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    f29(Ar_0, Ar_1, Ar_2) -> Com_1(f32(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 1:
	f26(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
	f27(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
	f29(Ar_0, Ar_1, Ar_2) -> Com_1(f32(Ar_0, Ar_1, Ar_2))
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(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 f20: X_2 - 1 >= 0 /\ -X_1 + X_2 + 4 >= 0 /\ -X_1 + 5 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] with all transitions in problem 4, the following new transition is obtained:
	koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 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, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
We thus obtain the following problem:
6:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ] with all transitions in problem 6, the following new transition is obtained:
	f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
We thus obtain the following problem:
7:	T:
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ] with all transitions in problem 7, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
We thus obtain the following problem:
8:	T:
		(Comp: ?, Cost: 2)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 8:
	f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
We thus obtain the following problem:
9:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f20) = 1
	Pol(f11) = 1
	Pol(f14) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
strictly and produces the following problem:
10:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ] with all transitions in problem 10, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
We thus obtain the following problem:
11:	T:
		(Comp: 1, Cost: 3)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ] with all transitions in problem 11, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 2, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 ]
We thus obtain the following problem:
12:	T:
		(Comp: 1, Cost: 5)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 2, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 2, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 ] with all transitions in problem 12, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 3, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 ]
We thus obtain the following problem:
13:	T:
		(Comp: 1, Cost: 7)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 3, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 3, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 ] with all transitions in problem 13, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 4, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 ]
We thus obtain the following problem:
14:	T:
		(Comp: 1, Cost: 9)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 4, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 4, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 ] with all transitions in problem 14, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 5, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 ]
We thus obtain the following problem:
15:	T:
		(Comp: 1, Cost: 11)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 5, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 5, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 ] with all transitions in problem 15, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 6, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 ]
We thus obtain the following problem:
16:	T:
		(Comp: 1, Cost: 13)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 6, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 6, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 ] with all transitions in problem 16, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 7, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 ]
We thus obtain the following problem:
17:	T:
		(Comp: 1, Cost: 15)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 7, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 7, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 ] with all transitions in problem 17, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 8, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 ]
We thus obtain the following problem:
18:	T:
		(Comp: 1, Cost: 17)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 8, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 8, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 ] with all transitions in problem 18, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 9, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 ]
We thus obtain the following problem:
19:	T:
		(Comp: 1, Cost: 19)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 9, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 9, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 ] with all transitions in problem 19, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 10, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 ]
We thus obtain the following problem:
20:	T:
		(Comp: 1, Cost: 21)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 10, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 10, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 ] with all transitions in problem 20, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 11, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 /\ Ar_0 + 11 >= 6 ]
We thus obtain the following problem:
21:	T:
		(Comp: 1, Cost: 23)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 11, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 /\ Ar_0 + 11 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 11, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 /\ Ar_0 + 11 >= 6 ] with all transitions in problem 21, the following new transition is obtained:
	f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 12, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 /\ Ar_0 + 11 >= 6 /\ Ar_0 + 12 >= 6 ]
We thus obtain the following problem:
22:	T:
		(Comp: 1, Cost: 25)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 12, Fresh_4', Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 /\ Ar_0 + 2 >= 6 /\ Ar_0 + 3 >= 6 /\ Ar_0 + 4 >= 6 /\ Ar_0 + 5 >= 6 /\ Ar_0 + 6 >= 6 /\ Ar_0 + 7 >= 6 /\ Ar_0 + 8 >= 6 /\ Ar_0 + 9 >= 6 /\ Ar_0 + 10 >= 6 /\ Ar_0 + 11 >= 6 /\ Ar_0 + 12 >= 6 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)     f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 2)     f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 /\ Ar_0 + 1 >= 6 ]
		(Comp: ?, Cost: 2)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)     f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.296 sec (SMT: 0.246 sec)
