
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 + 1, Ar_3)) [ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3))
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 + 1, Ar_3)) [ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 2 to obtain the following invariants:
  For symbol f1: -X_1 >= 0 /\ X_1 >= 0


This yielded the following problem:
3:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 + 1, Ar_3)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3))
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] with all transitions in problem 3, the following new transition is obtained:
	koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 + 1, Ar_3)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 4:
	f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3))
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= Ar_2 ]
		(Comp: 1, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 + 1, Ar_3)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.072 sec (SMT: 0.062 sec)
