WORST_CASE(?, O(1))

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_1, 0, 0, Ar_3, Ar_4)) [ 0 >= Fresh_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_0, 0, 0, Ar_3, Ar_4)) [ Fresh_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(0, 1023, 0, Ar_3, Ar_4))
		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Ar_0, Ar_1, Ar_2 + 1, Ar_3 + 2, Ar_4)) [ Ar_1 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ]
		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_1, Ar_2, Ar_4].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4))
		(Comp: ?, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4))
		(Comp: 1, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 1
	Pol(f0) = 1
	Pol(f7) = 1
	Pol(f21) = 0
orients all transitions weakly and the transitions
	f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ]
	f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ]
	f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ]
		(Comp: 1, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ]
		(Comp: 1, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4))
		(Comp: 1, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 1024
	Pol(f0) = 1024
	Pol(f7) = V_1 - V_2 + 1
	Pol(f21) = V_1 - V_2 + 1
orients all transitions weakly and the transition
	f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ]
strictly and produces the following problem:
5:	T:
		(Comp: 1, Cost: 0)       koat_start(Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)       f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ]
		(Comp: 1, Cost: 1)       f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ]
		(Comp: 1, Cost: 1)       f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ]
		(Comp: 1024, Cost: 1)    f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ]
		(Comp: 1, Cost: 1)       f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4))
		(Comp: 1, Cost: 1)       f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 1)       f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 1030

Time: 0.820 sec (SMT: 0.798 sec)