
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(1)) [ 0 >= Ar_0 /\ 3 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 1 produces the following problem:
2:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f4(1)) [ 0 >= Ar_0 /\ 3 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f4) = 1
	Pol(f12) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f4(1)) [ 0 >= Ar_0 /\ 3 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f4(Ar_0) -> Com_1(f4(1)) [ 0 >= Ar_0 /\ 3 >= Ar_0 ] with all transitions in problem 3, the following new transition is obtained:
	f4(Ar_0) -> Com_1(f4(2)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 2)    f4(Ar_0) -> Com_1(f4(2)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f4(Ar_0) -> Com_1(f4(2)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 ] with all transitions in problem 4, the following new transition is obtained:
	f4(Ar_0) -> Com_1(f4(3)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 ]
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 3)    f4(Ar_0) -> Com_1(f4(3)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f4(Ar_0) -> Com_1(f4(3)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 ] with all transitions in problem 5, the following new transition is obtained:
	f4(Ar_0) -> Com_1(f4(4)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 /\ 3 >= 3 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 4)    f4(Ar_0) -> Com_1(f4(4)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 /\ 3 >= 3 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f4(Ar_0) -> Com_1(f4(4)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 /\ 3 >= 3 ] with all transitions in problem 6, the following new transition is obtained:
	f4(Ar_0) -> Com_1(f12(4)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 /\ 3 >= 3 /\ 4 >= 4 ]
We thus obtain the following problem:
7:	T:
		(Comp: 1, Cost: 5)    f4(Ar_0) -> Com_1(f12(4)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 /\ 3 >= 3 /\ 4 >= 4 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] with all transitions in problem 7, the following new transition is obtained:
	koat_start(Ar_0) -> Com_1(f4(Fresh_0)) [ 0 <= 0 ]
We thus obtain the following problem:
8:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f4(Fresh_0)) [ 0 <= 0 ]
		(Comp: 1, Cost: 5)    f4(Ar_0) -> Com_1(f12(4)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 /\ 3 >= 3 /\ 4 >= 4 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f4(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 8:
	f0(Ar_0) -> Com_1(f4(Fresh_0))
We thus obtain the following problem:
9:	T:
		(Comp: 1, Cost: 5)    f4(Ar_0) -> Com_1(f12(4)) [ 0 >= Ar_0 /\ 3 >= Ar_0 /\ 3 >= 1 /\ 1 >= 1 /\ 3 >= 2 /\ 2 >= 1 /\ 3 >= 3 /\ 4 >= 4 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 4 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 3 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f4(Fresh_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.047 sec (SMT: 0.038 sec)
