MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(8, 0, 14, -1, Ar_4, Ar_5, Ar_6))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_3, Ar_5, Ar_6)) [ Ar_2 >= Ar_1 /\ Ar_0 >= H + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_2, Ar_5, Ar_6)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, Ar_1 - 1, Fresh_0, Fresh_1, Ar_5, Ar_6)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, Ar_4 - 1, Ar_3, Ar_4, Ar_5, Ar_6))
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_4 + 1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f20(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_3, Ar_3)) [ Ar_1 >= Ar_2 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 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_1, Ar_2, Ar_4].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f20(Ar_0, Ar_1, Ar_2, Ar_4)) [ Ar_1 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_4 + 1, Ar_2, Ar_4))
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_4 - 1, Ar_4))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_1 - 1, Fresh_1)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_2)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_3)) [ Ar_2 >= Ar_1 /\ Ar_0 >= H + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(8, 0, 14, Ar_4))
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f20(Ar_0, Ar_1, Ar_2, Ar_4)) [ Ar_1 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_4 + 1, Ar_2, Ar_4))
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_4 - 1, Ar_4))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_1 - 1, Fresh_1)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_2)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_3)) [ Ar_2 >= Ar_1 /\ Ar_0 >= H + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(8, 0, 14, Ar_4))
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 1
	Pol(f0) = 1
	Pol(f6) = 1
	Pol(f20) = 0
	Pol(f12) = 1
orients all transitions weakly and the transition
	f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f20(Ar_0, Ar_1, Ar_2, Ar_4)) [ Ar_1 >= Ar_2 + 1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f20(Ar_0, Ar_1, Ar_2, Ar_4)) [ Ar_1 >= Ar_2 + 1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_4 + 1, Ar_2, Ar_4))
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_4 - 1, Ar_4))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_1 - 1, Fresh_1)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_2)) [ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_3)) [ Ar_2 >= Ar_1 /\ Ar_0 >= H + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(8, 0, 14, Ar_4))
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f12: -X_2 + X_3 >= 0 /\ -X_1 + 8 >= 0 /\ X_1 - 8 >= 0
  For symbol f6: -X_1 + 8 >= 0 /\ X_1 - 8 >= 0


This yielded the following problem:
5:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(8, 0, 14, Ar_4))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_3)) [ -Ar_0 + 8 >= 0 /\ Ar_0 - 8 >= 0 /\ Ar_2 >= Ar_1 /\ Ar_0 >= H + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Fresh_2)) [ -Ar_0 + 8 >= 0 /\ Ar_0 - 8 >= 0 /\ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_1 - 1, Fresh_1)) [ -Ar_0 + 8 >= 0 /\ Ar_0 - 8 >= 0 /\ Ar_2 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_1, Ar_4 - 1, Ar_4)) [ -Ar_1 + Ar_2 >= 0 /\ -Ar_0 + 8 >= 0 /\ Ar_0 - 8 >= 0 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f6(Ar_0, Ar_4 + 1, Ar_2, Ar_4)) [ -Ar_1 + Ar_2 >= 0 /\ -Ar_0 + 8 >= 0 /\ Ar_0 - 8 >= 0 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f20(Ar_0, Ar_1, Ar_2, Ar_4)) [ -Ar_0 + 8 >= 0 /\ Ar_0 - 8 >= 0 /\ Ar_1 >= Ar_2 + 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 2.474 sec (SMT: 2.400 sec)