MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f14(Ar_0 - 1, Ar_1 - 1, Ar_2 + 1, Fresh_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 1 /\ Fresh_3 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f14(Ar_0 - 1, Ar_1, Ar_2, Fresh_2, Ar_4, Ar_5, Ar_6)) [ 0 >= Fresh_2 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6))
		(Comp: ?, Cost: 1)    f26(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f29(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f14(2*Fresh_0 + 1, Fresh_1, 0, Ar_3, Fresh_1, 2*Fresh_0 + 1, Fresh_0)) [ Fresh_1 >= 1 /\ 2*Fresh_0 >= 0 ]
		(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].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f14(2*Fresh_0 + 1, Fresh_1)) [ Fresh_1 >= 1 /\ 2*Fresh_0 >= 0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f26(Ar_0, Ar_1) -> Com_1(f29(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f24(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_2 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_3 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 2:
	f26(Ar_0, Ar_1) -> Com_1(f29(Ar_0, Ar_1))
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f24(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_3 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_2 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f14(2*Fresh_0 + 1, Fresh_1)) [ Fresh_1 >= 1 /\ 2*Fresh_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 3 produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f24(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_3 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_2 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f14(2*Fresh_0 + 1, Fresh_1)) [ Fresh_1 >= 1 /\ 2*Fresh_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(f24) = 0
	Pol(f14) = 1
	Pol(f0) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f14(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
strictly and produces the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f24(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_3 >= 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_2 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f14(2*Fresh_0 + 1, Fresh_1)) [ Fresh_1 >= 1 /\ 2*Fresh_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 5 to obtain the following invariants:
  For symbol f24: -X_1 >= 0


This yielded the following problem:
6:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f14(2*Fresh_0 + 1, Fresh_1)) [ Fresh_1 >= 1 /\ 2*Fresh_0 >= 0 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_2 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f14(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_3 >= 1 ]
		(Comp: 1, Cost: 1)    f14(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f24(Ar_0, Ar_1) -> Com_1(f24(Ar_0, Ar_1)) [ -Ar_0 >= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.513 sec (SMT: 1.462 sec)