
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_1)) [ Ar_0 >= 15 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_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 1:
	f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_1)) [ Ar_0 >= 15 /\ 0 >= Ar_0 ]
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ]
		(Comp: ?, Cost: 1)    f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f300(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)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f2) = 1
	Pol(f1) = 0
	Pol(f300) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transitions
	f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ]
	f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ]
strictly and produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f300(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(f300(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(f2(Ar_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(f2(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ]
		(Comp: 1, Cost: 1)    f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 5:
	f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1))
We thus obtain the following problem:
6:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.054 sec (SMT: 0.047 sec)
