
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(-Ar_0, Ar_1 + Ar_0, Ar_2 + Ar_0, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 + Ar_1, -Ar_1, Ar_2, Ar_3 + Ar_1, Ar_4, Ar_5)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1 + Ar_3, Ar_2, -Ar_3, Ar_4 + Ar_3, Ar_5)) [ 0 >= Ar_3 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2 + Ar_4, Ar_3 + Ar_4, -Ar_4, Ar_5)) [ 0 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0 + Ar_2, Ar_1, -Ar_2, Ar_3, Ar_4 + Ar_2, Ar_5)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4, Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2)) [ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_0, Ar_1, Ar_2, Ar_3, Ar_4].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4)) [ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_2, Ar_1, -Ar_2, Ar_3, Ar_4 + Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2 + Ar_4, Ar_3 + Ar_4, -Ar_4)) [ 0 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1 + Ar_3, Ar_2, -Ar_3, Ar_4 + Ar_3)) [ 0 >= Ar_3 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_1, -Ar_1, Ar_2, Ar_3 + Ar_1, Ar_4)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(-Ar_0, Ar_1 + Ar_0, Ar_2 + Ar_0, Ar_3, Ar_4)) [ 0 >= Ar_0 + 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_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4)) [ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_2, Ar_1, -Ar_2, Ar_3, Ar_4 + Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2 + Ar_4, Ar_3 + Ar_4, -Ar_4)) [ 0 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1 + Ar_3, Ar_2, -Ar_3, Ar_4 + Ar_3)) [ 0 >= Ar_3 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_1, -Ar_1, Ar_2, Ar_3 + Ar_1, Ar_4)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(-Ar_0, Ar_1 + Ar_0, Ar_2 + Ar_0, Ar_3, Ar_4)) [ 0 >= Ar_0 + 1 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] with all transitions in problem 3, the following new transition is obtained:
	koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4)) [ 0 <= 0 /\ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4)) [ 0 <= 0 /\ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4)) [ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_2, Ar_1, -Ar_2, Ar_3, Ar_4 + Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2 + Ar_4, Ar_3 + Ar_4, -Ar_4)) [ 0 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1 + Ar_3, Ar_2, -Ar_3, Ar_4 + Ar_3)) [ 0 >= Ar_3 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_1, -Ar_1, Ar_2, Ar_3 + Ar_1, Ar_4)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(-Ar_0, Ar_1 + Ar_0, Ar_2 + Ar_0, Ar_3, Ar_4)) [ 0 >= Ar_0 + 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 4:
	f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4)) [ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
We thus obtain the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(-Ar_0, Ar_1 + Ar_0, Ar_2 + Ar_0, Ar_3, Ar_4)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_1, -Ar_1, Ar_2, Ar_3 + Ar_1, Ar_4)) [ 0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1 + Ar_3, Ar_2, -Ar_3, Ar_4 + Ar_3)) [ 0 >= Ar_3 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2 + Ar_4, Ar_3 + Ar_4, -Ar_4)) [ 0 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0 + Ar_2, Ar_1, -Ar_2, Ar_3, Ar_4 + Ar_2)) [ 0 >= Ar_2 + 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4)) [ 0 <= 0 /\ Fresh_0 + Fresh_1 + Fresh_3 + Fresh_4 + Fresh_2 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.200 sec (SMT: 0.166 sec)
