
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(0, 99))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1)) [ Ar_1 >= Ar_0 + 1 /\ C >= D + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Fresh_0)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f11(Ar_0, Ar_1)) [ Ar_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 1 produces the following problem:
2:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(0, 99))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1)) [ Ar_1 >= Ar_0 + 1 /\ C >= D + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Fresh_0)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f11(Ar_0, Ar_1)) [ Ar_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

A polynomial rank function with
	Pol(f0) = 1
	Pol(f4) = 1
	Pol(f11) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f4(Ar_0, Ar_1) -> Com_1(f11(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(0, 99))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1)) [ Ar_1 >= Ar_0 + 1 /\ C >= D + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Fresh_0)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f11(Ar_0, Ar_1)) [ Ar_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

By chaining the transition koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] with all transitions in problem 3, the following new transition is obtained:
	koat_start(Ar_0, Ar_1) -> Com_1(f4(0, 99)) [ 0 <= 0 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f4(0, 99)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(0, 99))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1)) [ Ar_1 >= Ar_0 + 1 /\ C >= D + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Fresh_0)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f11(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 4:
	f0(Ar_0, Ar_1) -> Com_1(f4(0, 99))
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f11(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Fresh_0)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1)) [ Ar_1 >= Ar_0 + 1 /\ C >= D + 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f4(0, 99)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1) -> Com_1(f4(0, 99)) [ 0 <= 0 ] with all transitions in problem 5, the following new transitions are obtained:
	koat_start(Ar_0, Ar_1) -> Com_1(f4(0, Fresh_0)) [ 0 <= 0 /\ 99 >= 1 ]
	koat_start(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, 99)) [ 0 <= 0 /\ 99 >= 1 /\ C >= D + 1 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 2)    koat_start(Ar_0, Ar_1) -> Com_1(f4(0, Fresh_0)) [ 0 <= 0 /\ 99 >= 1 ]
		(Comp: 1, Cost: 2)    koat_start(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, 99)) [ 0 <= 0 /\ 99 >= 1 /\ C >= D + 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f11(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Fresh_0)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1)) [ Ar_1 >= Ar_0 + 1 /\ C >= D + 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.033 sec (SMT: 0.027 sec)
