
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1, Ar_2 + 1, Fresh_2, Ar_4, Ar_5)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(Ar_0 - 1, Ar_1, Ar_2, Fresh_1, Ar_4, Ar_5)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5))
		(Comp: ?, Cost: 1)    f23(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f26(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5))
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(4, Fresh_0, 0, Ar_3, Fresh_0, 4)) [ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(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].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f23(Ar_0, Ar_1) -> Com_1(f26(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 2:
	f23(Ar_0, Ar_1) -> Com_1(f26(Ar_0, Ar_1))
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
		(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 3 produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
		(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(f21) = 0
	Pol(f11) = 1
	Pol(f0) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
strictly and produces the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
		(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(f21) = V_1
	Pol(f11) = V_1
	Pol(f0) = 4
	Pol(koat_start) = 4
orients all transitions weakly and the transitions
	f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
	f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
strictly and produces the following problem:
6:	T:
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
		(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

Applied AI with 'oct' on problem 6 to obtain the following invariants:
  For symbol f11: -X_1 + X_2 + 3 >= 0 /\ -X_1 + 4 >= 0
  For symbol f21: -X_1 + X_2 + 3 >= 0 /\ -X_1 >= 0


This yielded the following problem:
7:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_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 7, the following new transition is obtained:
	koat_start(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ 0 <= 0 /\ Fresh_0 >= 1 ]
We thus obtain the following problem:
8:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ 0 <= 0 /\ Fresh_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 >= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 8:
	f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ]
We thus obtain the following problem:
9:	T:
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 >= 0 ]
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Ar_0 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ 0 <= 0 /\ Fresh_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Ar_0 ] with all transitions in problem 9, the following new transition is obtained:
	f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Ar_0 /\ -Ar_0 >= 0 ]
We thus obtain the following problem:
10:	T:
		(Comp: 1, Cost: 2)    f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Ar_0 /\ -Ar_0 >= 0 ]
		(Comp: ?, Cost: 1)    f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 >= 0 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ Ar_0 >= 1 /\ Fresh_2 >= 1 ]
		(Comp: 4, Cost: 1)    f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ 0 <= 0 /\ Fresh_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.070 sec (SMT: 0.058 sec)
