
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Ar_0 - 1, Ar_2, Ar_2 - 1, Ar_0, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3, Ar_2, Ar_0, Ar_6)) [ Ar_0 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Fresh_3)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Fresh_2, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f2(Fresh_0, Ar_1, Fresh_1, Ar_3, Ar_4, Ar_5, Ar_6))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 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_2].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f3(Ar_0, Ar_2)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1))
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2, Ar_2)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 2:
	f1(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 /\ Ar_2 >= 1 ]
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2, Ar_2)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f3(Ar_0, 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)    f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2, Ar_2)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f3(Ar_0, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f2) = 1
	Pol(f4) = 0
	Pol(f3) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
strictly and produces the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2, Ar_2)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f3(Ar_0, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2, Ar_2)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ] with all transitions in problem 5, the following new transition is obtained:
	f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
We thus obtain the following problem:
6:	T:
		(Comp: ?, Cost: 2)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f3(Ar_0, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_2) -> Com_1(f3(Ar_0, Ar_2)) [ 0 <= 0 ] with all transitions in problem 6, the following new transition is obtained:
	koat_start(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1)) [ 0 <= 0 ]
We thus obtain the following problem:
7:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 2)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 7:
	f3(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1))
We thus obtain the following problem:
8:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 2)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_2) -> Com_1(f2(Ar_0, Ar_2)) [ 0 >= Ar_0 ] with all transitions in problem 8, the following new transitions are obtained:
	f0(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
	f0(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_0 /\ 0 >= Ar_2 ]
We thus obtain the following problem:
9:	T:
		(Comp: ?, Cost: 3)    f0(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 2)    f0(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_0 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 2)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 9 produces the following problem:
10:	T:
		(Comp: ?, Cost: 3)    f0(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 2)    f0(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_0 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 2)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f4) = 0
	Pol(f2) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f0(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_0 /\ 0 >= Ar_2 ]
strictly and produces the following problem:
11:	T:
		(Comp: ?, Cost: 3)    f0(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 2)    f0(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_0 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 2)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_2) -> Com_1(f2(Fresh_0, Fresh_1)) [ 0 <= 0 ] with all transitions in problem 11, the following new transitions are obtained:
	koat_start(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Fresh_1 - 1)) [ 0 <= 0 /\ Fresh_2 >= 1 /\ Fresh_1 >= 1 ]
	koat_start(Ar_0, Ar_2) -> Com_1(f4(Fresh_0, Fresh_1)) [ 0 <= 0 /\ 0 >= Fresh_1 ]
We thus obtain the following problem:
12:	T:
		(Comp: 1, Cost: 3)    koat_start(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Fresh_1 - 1)) [ 0 <= 0 /\ Fresh_2 >= 1 /\ Fresh_1 >= 1 ]
		(Comp: 1, Cost: 2)    koat_start(Ar_0, Ar_2) -> Com_1(f4(Fresh_0, Fresh_1)) [ 0 <= 0 /\ 0 >= Fresh_1 ]
		(Comp: ?, Cost: 3)    f0(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 2)    f0(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_0 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 2)    f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 12:
	f2(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
	f2(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_2 ]
We thus obtain the following problem:
13:	T:
		(Comp: 1, Cost: 2)    f0(Ar_0, Ar_2) -> Com_1(f4(Ar_0, Ar_2)) [ 0 >= Ar_0 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 3)    f0(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Ar_2 - 1)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f0(Ar_0 - 1, Ar_2 - 1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 2)    koat_start(Ar_0, Ar_2) -> Com_1(f4(Fresh_0, Fresh_1)) [ 0 <= 0 /\ 0 >= Fresh_1 ]
		(Comp: 1, Cost: 3)    koat_start(Ar_0, Ar_2) -> Com_1(f0(Fresh_2 - 1, Fresh_1 - 1)) [ 0 <= 0 /\ Fresh_2 >= 1 /\ Fresh_1 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.090 sec (SMT: 0.075 sec)
