
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7) -> Com_1(f1(Ar_0, Ar_1, Fresh_8, Fresh_9, Fresh_10, Ar_5, Ar_6, Ar_7)) [ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7) -> Com_1(f1(Ar_0, Ar_1, Fresh_5, Fresh_6, Fresh_7, Ar_5, Ar_6, Ar_7)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7) -> Com_1(f300(Ar_0, Ar_0, Fresh_2, Fresh_3, Ar_4, Fresh_4, Ar_6, Ar_7)) [ Ar_0 = Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Fresh_0, Fresh_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7)) [ 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].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    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(f1(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= Ar_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_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ 0 <= 0 ] with all transitions in problem 3, the following new transition is obtained:
	koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 4:
	f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1))
We thus obtain the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.025 sec (SMT: 0.019 sec)
