MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1200(Ar_1, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f2200(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10000(Ar_0, Ar_1, 0, Ar_3)) [ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f100(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f110(Ar_0, 1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f110(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f120(Ar_0, 2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f120(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f120(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f1200(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1200(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, 2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, 1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f110(1, 1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f12(Ar_1, 2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_1, Ar_1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_1, 1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f100(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f110(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f120(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f1000(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1200(Ar_0, 2, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f1000(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f1200(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, Ar_1, Ar_2, 1))
		(Comp: ?, Cost: 1)    f1000(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10001(Ar_0, 1, Ar_2, 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

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_2].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1000(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f1200(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f1000(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f1000(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: ?, Cost: 1)    f120(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f110(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f100(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f12(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f110(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: ?, Cost: 1)    f1200(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: ?, Cost: 1)    f120(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: ?, Cost: 1)    f110(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: ?, Cost: 1)    f100(Ar_2) -> Com_1(f110(Ar_2))
		(Comp: ?, Cost: 1)    f12(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: ?, Cost: 1)    f2200(Ar_2) -> Com_1(f10000(0)) [ Ar_2 = 0 ]
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f1000(Ar_2) -> Com_1(f10001(Ar_2))
	f1000(Ar_2) -> Com_1(f10001(Ar_2))
	f1000(Ar_2) -> Com_1(f1200(Ar_2))
	f100(Ar_2) -> Com_1(f10001(Ar_2))
	f100(Ar_2) -> Com_1(f110(Ar_2))
	f2200(Ar_2) -> Com_1(f10000(0)) [ Ar_2 = 0 ]
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f1200(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: ?, Cost: 1)    f1200(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f120(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: ?, Cost: 1)    f120(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: ?, Cost: 1)    f110(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: ?, Cost: 1)    f110(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f12(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: ?, Cost: 1)    f12(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f110(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 3 produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f1200(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: ?, Cost: 1)    f1200(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f120(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: ?, Cost: 1)    f120(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: 1, Cost: 1)    f110(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: 1, Cost: 1)    f110(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f12(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: ?, Cost: 1)    f12(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f110(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f1200) = 1
	Pol(f10001) = 0
	Pol(f120) = 1
	Pol(f2) = 2
	Pol(f110) = 1
	Pol(f12) = 1
	Pol(f0) = 2
	Pol(koat_start) = 2
orients all transitions weakly and the transitions
	f2(Ar_2) -> Com_1(f1200(Ar_2))
	f2(Ar_2) -> Com_1(f10001(Ar_2))
	f1200(Ar_2) -> Com_1(f10001(Ar_2))
	f120(Ar_2) -> Com_1(f10001(Ar_2))
	f12(Ar_2) -> Com_1(f10001(Ar_2))
strictly and produces the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f1200(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: 2, Cost: 1)    f1200(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f120(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: 2, Cost: 1)    f120(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f2(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: 2, Cost: 1)    f2(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 2, Cost: 1)    f2(Ar_2) -> Com_1(f1200(Ar_2))
		(Comp: 1, Cost: 1)    f110(Ar_2) -> Com_1(f120(Ar_2))
		(Comp: 1, Cost: 1)    f110(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: ?, Cost: 1)    f12(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: 2, Cost: 1)    f12(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f2(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f110(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f12(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 1)    f0(Ar_2) -> Com_1(f10001(Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.160 sec (SMT: 1.102 sec)