
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0, Fresh_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_0 + 2 /\ Fresh_1^2 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Fresh_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 + 2 /\ Ar_2 >= Fresh_0^2 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 + 1 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0, Fresh_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_0 + 2 /\ Fresh_1^2 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Fresh_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 + 2 /\ Ar_2 >= Fresh_0^2 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 + 1 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f6) = 1
	Pol(f16) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 + 1 >= Ar_1 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0, Fresh_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_0 + 2 /\ Fresh_1^2 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Fresh_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 + 2 /\ Ar_2 >= Fresh_0^2 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 + 1 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] with all transitions in problem 3, the following new transition is obtained:
	koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3)) [ 0 <= 0 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0, Fresh_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_0 + 2 /\ Fresh_1^2 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Fresh_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 + 2 /\ Ar_2 >= Fresh_0^2 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 + 1 >= 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, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3))
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 + 1 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Fresh_0, Ar_1, Ar_2, Fresh_0)) [ Ar_1 >= Ar_0 + 2 /\ Ar_2 >= Fresh_0^2 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0, Fresh_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_0 + 2 /\ Fresh_1^2 >= Ar_2 + 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(1, Ar_2, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.031 sec (SMT: 0.023 sec)
