MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(Fresh_3, 0))
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_2, Ar_1 + 1)) [ 3 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1 + 1)) [ Ar_1 >= 5 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, 1)) [ Ar_1 = 4 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f12(0, Ar_1)) [ Ar_0 = 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 1 produces the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(Fresh_3, 0))
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_2, Ar_1 + 1)) [ 3 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1 + 1)) [ Ar_1 >= 5 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, 1)) [ Ar_1 = 4 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f12(0, Ar_1)) [ Ar_0 = 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f4) = 1
	Pol(f5) = 1
	Pol(f0) = 1
	Pol(f12) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f4(Ar_0, Ar_1) -> Com_1(f12(0, Ar_1)) [ Ar_0 = 0 ]
strictly and produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(Fresh_3, 0))
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_2, Ar_1 + 1)) [ 3 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1 + 1)) [ Ar_1 >= 5 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, 1)) [ Ar_1 = 4 ]
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f12(0, Ar_1)) [ Ar_0 = 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f4: X_2 >= 0
  For symbol f5: X_2 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f12(0, Ar_1)) [ Ar_1 >= 0 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, 1)) [ Ar_1 >= 0 /\ Ar_1 = 4 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_1, Ar_1 + 1)) [ Ar_1 >= 0 /\ Ar_1 >= 5 ]
		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1) -> Com_1(f4(Fresh_2, Ar_1 + 1)) [ Ar_1 >= 0 /\ 3 >= Ar_1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(Fresh_3, 0))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ Ar_1 >= 0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f5(Ar_0, Ar_1)) [ Ar_1 >= 0 /\ 0 >= Ar_0 + 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.498 sec (SMT: 1.446 sec)