
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f3(0, Ar_1))
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 1 produces the following problem:
2:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f3(0, Ar_1))
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f3) = 1
	Pol(f12) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f3(0, Ar_1))
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f3(0, Ar_1)) with all transitions in problem 3, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f3(1, 0)) [ 9 >= 0 /\ 4 >= 0 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 2)    f0(Ar_0, Ar_1) -> Com_1(f3(1, 0)) [ 9 >= 0 /\ 4 >= 0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f3(1, 0)) [ 9 >= 0 /\ 4 >= 0 ] with all transitions in problem 4, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f3(2, 1)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 ]
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 3)    f0(Ar_0, Ar_1) -> Com_1(f3(2, 1)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f3(2, 1)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 ] with all transitions in problem 5, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f3(3, 2)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 4)    f0(Ar_0, Ar_1) -> Com_1(f3(3, 2)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f3(3, 2)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 ] with all transitions in problem 6, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f3(4, 3)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 ]
We thus obtain the following problem:
7:	T:
		(Comp: 1, Cost: 5)    f0(Ar_0, Ar_1) -> Com_1(f3(4, 3)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f3(4, 3)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 ] with all transitions in problem 7, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f3(5, 4)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 ]
We thus obtain the following problem:
8:	T:
		(Comp: 1, Cost: 6)    f0(Ar_0, Ar_1) -> Com_1(f3(5, 4)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f3(5, 4)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 ] with all transitions in problem 8, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, 4)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 /\ 5 = 5 ]
We thus obtain the following problem:
9:	T:
		(Comp: 1, Cost: 7)    f0(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, 4)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 /\ 5 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] with all transitions in problem 9, the following new transition is obtained:
	koat_start(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, 4)) [ 0 <= 0 /\ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 /\ 5 = 5 ]
We thus obtain the following problem:
10:	T:
		(Comp: 1, Cost: 7)    koat_start(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, 4)) [ 0 <= 0 /\ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 /\ 5 = 5 ]
		(Comp: 1, Cost: 7)    f0(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, 4)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 /\ 5 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 10:
	f0(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, 4)) [ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 /\ 5 = 5 ]
We thus obtain the following problem:
11:	T:
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f12(Ar_0, Ar_1)) [ Ar_0 >= 10 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Ar_0 + 1, Ar_0)) [ 9 >= Ar_0 /\ 4 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, Ar_1)) [ Ar_0 = 5 ]
		(Comp: 1, Cost: 7)    koat_start(Ar_0, Ar_1) -> Com_1(f3(Fresh_0 + 1, 4)) [ 0 <= 0 /\ 9 >= 0 /\ 4 >= 0 /\ 9 >= 1 /\ 4 >= 1 /\ 9 >= 2 /\ 4 >= 2 /\ 9 >= 3 /\ 4 >= 3 /\ 9 >= 4 /\ 4 >= 4 /\ 5 = 5 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.085 sec (SMT: 0.069 sec)
