MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f12(Fresh_3, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ 15 >= L ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f20(Ar_0, Fresh_2, Fresh_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f20(Ar_0, Fresh_1, Fresh_1, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f5(Ar_0, Ar_1, Ar_2, 1, 1, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ 15 >= M ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f12(Ar_0 - 1, Ar_1, Ar_2, Ar_3, Ar_4, 0, 0, Ar_7, Ar_8, Ar_9, Ar_10))
		(Comp: ?, Cost: 1)    f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f5(0, Ar_1, Ar_2, 0, 0, Ar_5, Ar_6, Fresh_0, Fresh_0, 0, 0)) [ Ar_0 = 0 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ 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].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f5(0)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0) -> Com_1(f12(Ar_0 - 1))
		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f5(Ar_0)) [ 15 >= M ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f20(Ar_0)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f20(Ar_0)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_3)) [ 15 >= L ]
		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f5(Ar_0))
	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) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f5(0)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0) -> Com_1(f12(Ar_0 - 1))
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f5(Ar_0)) [ 15 >= M ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f20(Ar_0)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f20(Ar_0)) [ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_3)) [ 15 >= L ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f5(Ar_0))
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(koat_start) = 1
	Pol(f0) = 1
	Pol(f12) = 1
	Pol(f5) = 0
	Pol(f20) = 1
orients all transitions weakly and the transition
	f12(Ar_0) -> Com_1(f5(0)) [ Ar_0 = 0 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f12(Ar_0) -> Com_1(f5(0)) [ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f20(Ar_0) -> Com_1(f12(Ar_0 - 1))
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f5(Ar_0)) [ 15 >= M ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f20(Ar_0)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f12(Ar_0) -> Com_1(f20(Ar_0)) [ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_3)) [ 15 >= L ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f5(Ar_0))
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.614 sec (SMT: 0.582 sec)