WORST_CASE(?, O(1))

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f54(Ar_0, Ar_1, Ar_1 + Ar_0, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ Ar_0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f54(Ar_0, Ar_1, Ar_1 - Ar_0, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f38(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f11(Fresh_3, Ar_1, Ar_2, Ar_3, Ar_4, Ar_3, Ar_3, Ar_7, Ar_8, Ar_9, Ar_10)) [ Ar_3 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f38(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f11(Fresh_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_3, Ar_3, Fresh_2, Ar_8, Ar_9, Ar_10)) [ Ar_4 >= Ar_3 ]
		(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(f38(1, 2, Ar_2, 1, 10, Ar_5, Ar_6, Ar_7, 10, 2, Fresh_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, Ar_1, Ar_3, Ar_4].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_3, Ar_4)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f38(1, 2, 1, 10))
		(Comp: ?, Cost: 1)    f38(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f11(Fresh_1, Ar_1, Ar_3, Ar_4)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f38(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f11(Fresh_3, Ar_1, Ar_3, Ar_4)) [ Ar_3 >= Ar_4 + 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_3, Ar_4)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_3, Ar_4)) [ Ar_0 >= Ar_1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 2:
	f38(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f11(Fresh_3, Ar_1, Ar_3, Ar_4)) [ Ar_3 >= Ar_4 + 1 ]
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_3, Ar_4)) [ Ar_0 >= Ar_1 ]
		(Comp: ?, Cost: 1)    f11(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_3, Ar_4)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f38(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f11(Fresh_1, Ar_1, Ar_3, Ar_4)) [ Ar_4 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f38(1, 2, 1, 10))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_3, Ar_4)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 3 produces the following problem:
4:	T:
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_3, Ar_4)) [ Ar_0 >= Ar_1 ]
		(Comp: 1, Cost: 1)    f11(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_3, Ar_4)) [ Ar_1 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)    f38(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f11(Fresh_1, Ar_1, Ar_3, Ar_4)) [ Ar_4 >= Ar_3 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f38(1, 2, 1, 10))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_3, Ar_4)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 4

Time: 0.312 sec (SMT: 0.304 sec)