MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f16(Ar_0, 0, Fresh_2, Ar_3, Ar_4)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))
		(Comp: ?, Cost: 1)    f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f30(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f10(Fresh_1, Ar_1, Ar_2, 0, Fresh_1)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f10(Fresh_0, 0, Ar_2, 0, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 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_2].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f0(Ar_0, Ar_2)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f10(Fresh_0, Ar_2))
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_2) -> Com_1(f10(Fresh_1, Ar_2)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f27(Ar_0, Ar_2) -> Com_1(f30(Ar_0, Ar_2))
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2))
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f16(Ar_0, Ar_2) -> Com_1(f10(Fresh_1, Ar_2)) [ 0 >= Ar_2 ]
	f27(Ar_0, Ar_2) -> Com_1(f30(Ar_0, Ar_2))
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2))
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f10(Fresh_0, Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f0(Ar_0, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 3 produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Ar_2)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2))
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 ]
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f10(Fresh_0, Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f0(Ar_0, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f16: X_2 - 1 >= 0 /\ -X_1 + X_2 - 1 >= 0 /\ -X_1 >= 0
  For symbol f25: X_1 - 1 >= 0


This yielded the following problem:
5:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_2) -> Com_1(f0(Ar_0, Ar_2)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_2) -> Com_1(f10(Fresh_0, Ar_2))
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 /\ Fresh_2 >= 1 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_2) -> Com_1(f25(Ar_0, Ar_2)) [ Ar_0 - 1 >= 0 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_2) -> Com_1(f16(Ar_0, Ar_2)) [ Ar_2 - 1 >= 0 /\ -Ar_0 + Ar_2 - 1 >= 0 /\ -Ar_0 >= 0 /\ Ar_2 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.992 sec (SMT: 0.950 sec)