MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f26(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    f27(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    f29(Ar_0, Ar_1, Ar_2) -> Com_1(f32(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 1:
	f26(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
	f27(Ar_0, Ar_1, Ar_2) -> Com_1(f27(Ar_0, Ar_1, Ar_2))
	f29(Ar_0, Ar_1, Ar_2) -> Com_1(f32(Ar_0, Ar_1, Ar_2))
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ 2 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f20: X_2 - 1 >= 0 /\ -X_1 + X_2 + 4 >= 0 /\ -X_1 + 5 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Fresh_0, Fresh_1, Fresh_0))
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ 5 >= Ar_0 /\ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_4, Ar_2)) [ Ar_0 >= 6 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0 + 1, Fresh_2, Ar_2)) [ 5 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ Ar_0 >= 3 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Fresh_3, Ar_2)) [ Ar_1 - 1 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 5 >= 0 /\ 2 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 2.068 sec (SMT: 2.006 sec)