
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 + 1, Fresh_11, Fresh_11, Ar_3, Ar_4, Ar_5)) [ Fresh_11 >= 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 + 1, Fresh_10, Fresh_10, Ar_3, Ar_4, Ar_5)) [ 0 >= Fresh_10 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, 0, 0, Fresh_9, Ar_4, Ar_5))
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 + 1, Fresh_8, Fresh_8, Ar_3, Ar_4, Ar_5)) [ Fresh_8 >= 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 + 1, Fresh_7, Fresh_7, Ar_3, Ar_4, Ar_5)) [ 0 >= Fresh_7 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, 0, 0, Fresh_6, Ar_4, Ar_5))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Fresh_4 + 1, Fresh_5, Fresh_5, Ar_3, Fresh_4, Fresh_4)) [ Fresh_5 >= 1 /\ Fresh_4 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Fresh_2 + 1, Fresh_3, Fresh_3, Ar_3, Fresh_2, Fresh_2)) [ 0 >= Fresh_3 + 1 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Fresh_0, 0, 0, Fresh_1, Fresh_0, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(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].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f4(Ar_0)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f0(Fresh_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_2 + 1)) [ 0 >= Fresh_3 + 1 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_4 + 1)) [ Fresh_5 >= 1 /\ Fresh_4 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f0(Ar_0))
		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ 0 >= Fresh_7 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Fresh_8 >= 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f0(Ar_0))
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ 0 >= Fresh_10 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Fresh_11 >= 1 /\ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f3(Ar_0) -> Com_1(f0(Ar_0))
	f3(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ 0 >= Fresh_7 + 1 /\ Ar_0 >= 1 ]
	f3(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Fresh_8 >= 1 /\ Ar_0 >= 1 ]
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Fresh_11 >= 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ 0 >= Fresh_10 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f0(Ar_0))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_4 + 1)) [ Fresh_5 >= 1 /\ Fresh_4 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_2 + 1)) [ 0 >= Fresh_3 + 1 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f0(Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f4(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 3 produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Fresh_11 >= 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ 0 >= Fresh_10 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f0(Ar_0))
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_4 + 1)) [ Fresh_5 >= 1 /\ Fresh_4 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_2 + 1)) [ 0 >= Fresh_3 + 1 /\ Fresh_2 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f0(Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f4(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f2) = 1
	Pol(f0) = 0
	Pol(f4) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f2(Ar_0) -> Com_1(f0(Ar_0))
strictly and produces the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Fresh_11 >= 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ 0 >= Fresh_10 + 1 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0) -> Com_1(f0(Ar_0))
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_4 + 1)) [ Fresh_5 >= 1 /\ Fresh_4 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_2 + 1)) [ 0 >= Fresh_3 + 1 /\ Fresh_2 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f0(Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f4(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 5 to obtain the following invariants:
  For symbol f2: X_1 - 2 >= 0


This yielded the following problem:
6:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f4(Ar_0)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f0(Fresh_0))
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_2 + 1)) [ 0 >= Fresh_3 + 1 /\ Fresh_2 >= 1 ]
		(Comp: 1, Cost: 1)    f4(Ar_0) -> Com_1(f2(Fresh_4 + 1)) [ Fresh_5 >= 1 /\ Fresh_4 >= 1 ]
		(Comp: 1, Cost: 1)    f2(Ar_0) -> Com_1(f0(Ar_0)) [ Ar_0 - 2 >= 0 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Ar_0 - 2 >= 0 /\ 0 >= Fresh_10 + 1 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0) -> Com_1(f2(Ar_0 + 1)) [ Ar_0 - 2 >= 0 /\ Fresh_11 >= 1 /\ Ar_0 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.041 sec (SMT: 0.035 sec)
