MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f8(1, 1, 0, 1, 1, Ar_5))
		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 29 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f14(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_0, Ar_5)) [ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f14(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4 + 2, Ar_5)) [ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3 + 10, Ar_4, Ar_5)) [ 12 >= Ar_4 /\ Ar_4 >= 10 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4, Ar_5)) [ Ar_4 >= 13 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4, Ar_5)) [ 9 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f8(Ar_0, Ar_1, Ar_2, Ar_3 + 2, Ar_4 - 10, Ar_5)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f28(Ar_0, Ar_1, 1, Ar_3, Ar_4, 1)) [ Ar_3 >= 30 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_3, Ar_4].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\ Ar_4 >= 10 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
		(Comp: ?, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))
	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_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\ Ar_4 >= 10 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
		(Comp: ?, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]
		(Comp: 1, Cost: 1)    f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 1
	Pol(f0) = 1
	Pol(f8) = 1
	Pol(f28) = 0
	Pol(f10) = 1
	Pol(f14) = 1
orients all transitions weakly and the transition
	f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]
		(Comp: ?, Cost: 1)    f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\ Ar_4 >= 10 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
		(Comp: ?, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
		(Comp: ?, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]
		(Comp: 1, Cost: 1)    f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 230
	Pol(f0) = 230
	Pol(f8) = -8*V_1 + 238
	Pol(f28) = -8*V_1 + 238
	Pol(f10) = -8*V_1 + 222
	Pol(f14) = -8*V_1 + 222
orients all transitions weakly and the transition
	f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]
strictly and produces the following problem:
5:	T:
		(Comp: 1, Cost: 0)      koat_start(Ar_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)      f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]
		(Comp: ?, Cost: 1)      f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\ Ar_4 >= 10 ]
		(Comp: ?, Cost: 1)      f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
		(Comp: ?, Cost: 1)      f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
		(Comp: 230, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]
		(Comp: 1, Cost: 1)      f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f14) = 1
	Pol(f10) = 1
	Pol(f8) = 0
and size complexities
	S("f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))", 0-0) = 1
	S("f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))", 0-1) = 1
	S("f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]", 0-0) = ?
	S("f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]", 0-1) = ?
	S("f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\\ Ar_4 >= 6 ]", 0-0) = ?
	S("f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\\ Ar_4 >= 6 ]", 0-1) = ?
	S("f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\\ 5 >= Ar_4 ]", 0-0) = ?
	S("f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\\ 5 >= Ar_4 ]", 0-1) = ?
	S("f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\\ Ar_4 >= 10 ]", 0-0) = ?
	S("f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\\ Ar_4 >= 10 ]", 0-1) = 12
	S("f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]", 0-0) = ?
	S("f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]", 0-1) = ?
	S("f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]", 0-0) = ?
	S("f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]", 0-1) = ?
	S("f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]", 0-0) = ?
	S("f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]", 0-1) = ?
	S("f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]", 0-0) = ?
	S("f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]", 0-1) = ?
	S("koat_start(Ar_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]", 0-0) = Ar_3
	S("koat_start(Ar_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]", 0-1) = Ar_4
orients the transitions
	f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\ Ar_4 >= 10 ]
	f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]
	f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]
	f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]
	f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
	f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
weakly and the transition
	f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]
strictly and produces the following problem:
6:	T:
		(Comp: 1, Cost: 0)      koat_start(Ar_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)      f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 >= 30 ]
		(Comp: 230, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ 9 >= Ar_4 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_4 >= 13 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ 12 >= Ar_4 /\ Ar_4 >= 10 ]
		(Comp: ?, Cost: 1)      f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
		(Comp: ?, Cost: 1)      f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
		(Comp: 230, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ 29 >= Ar_3 ]
		(Comp: 1, Cost: 1)      f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 6 to obtain the following invariants:
  For symbol f10: X_1 - 1 >= 0
  For symbol f14: X_1 - 1 >= 0
  For symbol f8: X_1 - 1 >= 0


This yielded the following problem:
7:	T:
		(Comp: 1, Cost: 1)      f0(Ar_3, Ar_4) -> Com_1(f8(1, 1))
		(Comp: 230, Cost: 1)    f8(Ar_3, Ar_4) -> Com_1(f10(Ar_3, Ar_4)) [ Ar_3 - 1 >= 0 /\ 29 >= Ar_3 ]
		(Comp: ?, Cost: 1)      f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Fresh_0)) [ Ar_3 - 1 >= 0 /\ Ar_3 >= Ar_4 + 1 /\ Ar_4 >= 6 ]
		(Comp: ?, Cost: 1)      f10(Ar_3, Ar_4) -> Com_1(f14(Ar_3, Ar_4 + 2)) [ Ar_3 - 1 >= 0 /\ Ar_3 >= Ar_4 + 1 /\ 5 >= Ar_4 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 10, Ar_4)) [ Ar_3 - 1 >= 0 /\ 12 >= Ar_4 /\ Ar_4 >= 10 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_3 - 1 >= 0 /\ Ar_4 >= 13 ]
		(Comp: ?, Cost: 1)      f14(Ar_3, Ar_4) -> Com_1(f10(Ar_3 + 1, Ar_4)) [ Ar_3 - 1 >= 0 /\ 9 >= Ar_4 ]
		(Comp: 230, Cost: 1)    f10(Ar_3, Ar_4) -> Com_1(f8(Ar_3 + 2, Ar_4 - 10)) [ Ar_3 - 1 >= 0 /\ Ar_4 >= Ar_3 ]
		(Comp: 1, Cost: 1)      f8(Ar_3, Ar_4) -> Com_1(f28(Ar_3, Ar_4)) [ Ar_3 - 1 >= 0 /\ Ar_3 >= 30 ]
		(Comp: 1, Cost: 0)      koat_start(Ar_3, Ar_4) -> Com_1(f0(Ar_3, Ar_4)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 2.813 sec (SMT: 2.732 sec)