MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, 0, Fresh_2, Ar_3)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f22(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f24(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f27(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_1, Ar_1, Ar_2, 0)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_0, 0, Ar_2, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 1:
	f24(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f27(Ar_0, Ar_1, Ar_2, Ar_3))
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f22(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_1, Ar_1, Ar_2, 0)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, 0, Fresh_2, Ar_3)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_0, 0, Ar_2, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f22(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_1, Ar_1, Ar_2, 0)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, 0, Fresh_2, Ar_3)) [ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_0, 0, Ar_2, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f22) = 0
	Pol(f14) = 1
	Pol(f9) = 1
	Pol(f0) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 1 ]
strictly and produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f22(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_1, Ar_1, Ar_2, 0)) [ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_2 >= 1 ]
		(Comp: 1, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, 0, Fresh_2, Ar_3)) [ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_0, 0, Ar_2, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f14: -X_4 >= 0 /\ X_2 - X_4 >= 0 /\ -X_2 - X_4 >= 0 /\ -X_1 - X_4 >= 0 /\ X_4 >= 0 /\ X_2 + X_4 >= 0 /\ -X_2 + X_4 >= 0 /\ -X_1 + X_4 >= 0 /\ -X_2 >= 0 /\ -X_1 - X_2 >= 0 /\ X_2 >= 0 /\ -X_1 + X_2 >= 0 /\ -X_1 >= 0
  For symbol f22: -X_4 >= 0 /\ X_2 - X_4 >= 0 /\ -X_2 - X_4 >= 0 /\ X_1 - X_4 - 1 >= 0 /\ X_4 >= 0 /\ X_2 + X_4 >= 0 /\ -X_2 + X_4 >= 0 /\ X_1 + X_4 - 1 >= 0 /\ -X_2 >= 0 /\ X_1 - X_2 - 1 >= 0 /\ X_2 >= 0 /\ X_1 + X_2 - 1 >= 0 /\ X_1 - 1 >= 0
  For symbol f9: -X_4 >= 0 /\ X_2 - X_4 >= 0 /\ -X_2 - X_4 >= 0 /\ X_4 >= 0 /\ X_2 + X_4 >= 0 /\ -X_2 + X_4 >= 0 /\ -X_2 >= 0 /\ X_2 >= 0


This yielded the following problem:
5:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_0, 0, Ar_2, 0))
		(Comp: ?, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, 0, Fresh_2, Ar_3)) [ -Ar_3 >= 0 /\ Ar_1 - Ar_3 >= 0 /\ -Ar_1 - Ar_3 >= 0 /\ Ar_3 >= 0 /\ Ar_1 + Ar_3 >= 0 /\ -Ar_1 + Ar_3 >= 0 /\ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f9(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ -Ar_3 >= 0 /\ Ar_1 - Ar_3 >= 0 /\ -Ar_1 - Ar_3 >= 0 /\ Ar_3 >= 0 /\ Ar_1 + Ar_3 >= 0 /\ -Ar_1 + Ar_3 >= 0 /\ -Ar_1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1, Ar_3)) [ -Ar_3 >= 0 /\ Ar_1 - Ar_3 >= 0 /\ -Ar_1 - Ar_3 >= 0 /\ -Ar_0 - Ar_3 >= 0 /\ Ar_3 >= 0 /\ Ar_1 + Ar_3 >= 0 /\ -Ar_1 + Ar_3 >= 0 /\ -Ar_0 + Ar_3 >= 0 /\ -Ar_1 >= 0 /\ -Ar_0 - Ar_1 >= 0 /\ Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ -Ar_0 >= 0 /\ Ar_2 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f9(Fresh_1, Ar_1, Ar_2, 0)) [ -Ar_3 >= 0 /\ Ar_1 - Ar_3 >= 0 /\ -Ar_1 - Ar_3 >= 0 /\ -Ar_0 - Ar_3 >= 0 /\ Ar_3 >= 0 /\ Ar_1 + Ar_3 >= 0 /\ -Ar_1 + Ar_3 >= 0 /\ -Ar_0 + Ar_3 >= 0 /\ -Ar_1 >= 0 /\ -Ar_0 - Ar_1 >= 0 /\ Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ -Ar_0 >= 0 /\ 0 >= Ar_2 ]
		(Comp: ?, Cost: 1)    f22(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ -Ar_3 >= 0 /\ Ar_1 - Ar_3 >= 0 /\ -Ar_1 - Ar_3 >= 0 /\ Ar_0 - Ar_3 - 1 >= 0 /\ Ar_3 >= 0 /\ Ar_1 + Ar_3 >= 0 /\ -Ar_1 + Ar_3 >= 0 /\ Ar_0 + Ar_3 - 1 >= 0 /\ -Ar_1 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 - 1 >= 0 /\ Ar_0 - 1 >= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 2.607 sec (SMT: 2.520 sec)