
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f1(Ar_0) -> Com_1(f4(3))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f1(Ar_0)) [ 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)    f1(Ar_0) -> Com_1(f4(3))
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f1(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 2 to obtain the following invariants:
  For symbol f4: -X_1 + 3 >= 0 /\ X_1 - 3 >= 0


This yielded the following problem:
3:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f1(Ar_0)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(3)) [ -Ar_0 + 3 >= 0 /\ Ar_0 - 3 >= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0) -> Com_1(f4(3))
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0) -> Com_1(f1(Ar_0)) [ 0 <= 0 ] with all transitions in problem 3, the following new transition is obtained:
	koat_start(Ar_0) -> Com_1(f4(3)) [ 0 <= 0 ]
We thus obtain the following problem:
4:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f4(3)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(3)) [ -Ar_0 + 3 >= 0 /\ Ar_0 - 3 >= 0 ]
		(Comp: 1, Cost: 1)    f1(Ar_0) -> Com_1(f4(3))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 4:
	f1(Ar_0) -> Com_1(f4(3))
We thus obtain the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(3)) [ -Ar_0 + 3 >= 0 /\ Ar_0 - 3 >= 0 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f4(3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0) -> Com_1(f4(3)) [ 0 <= 0 ] with all transitions in problem 5, the following new transition is obtained:
	koat_start(Ar_0) -> Com_1(f4(3)) [ 0 <= 0 /\ 0 >= 0 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 2)    koat_start(Ar_0) -> Com_1(f4(3)) [ 0 <= 0 /\ 0 >= 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f4(3)) [ -Ar_0 + 3 >= 0 /\ Ar_0 - 3 >= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.017 sec (SMT: 0.015 sec)
