WORST_CASE(?, O(n^1))

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    eval_abc_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb0_in(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    eval_abc_bb0_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_0(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    eval_abc_0(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_1(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    eval_abc_1(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_2(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    eval_abc_2(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_3(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    eval_abc_3(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_4(Ar_0, Ar_1, Ar_2))
		(Comp: ?, Cost: 1)    eval_abc_4(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_0))
		(Comp: ?, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb2_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 <= Ar_1 ]
		(Comp: ?, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb3_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 > Ar_1 ]
		(Comp: ?, Cost: 1)    eval_abc_bb2_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_2 + 1))
		(Comp: ?, Cost: 1)    eval_abc_bb3_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_stop(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 1 produces the following problem:
2:	T:
		(Comp: 1, Cost: 1)    eval_abc_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb0_in(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_bb0_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_0(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_0(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_1(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_1(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_2(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_2(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_3(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_3(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_4(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_4(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_0))
		(Comp: ?, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb2_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 <= Ar_1 ]
		(Comp: ?, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb3_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 > Ar_1 ]
		(Comp: ?, Cost: 1)    eval_abc_bb2_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_2 + 1))
		(Comp: ?, Cost: 1)    eval_abc_bb3_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_stop(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(eval_abc_start) = 2
	Pol(eval_abc_bb0_in) = 2
	Pol(eval_abc_0) = 2
	Pol(eval_abc_1) = 2
	Pol(eval_abc_2) = 2
	Pol(eval_abc_3) = 2
	Pol(eval_abc_4) = 2
	Pol(eval_abc_bb1_in) = 2
	Pol(eval_abc_bb2_in) = 2
	Pol(eval_abc_bb3_in) = 1
	Pol(eval_abc_stop) = 0
	Pol(koat_start) = 2
orients all transitions weakly and the transitions
	eval_abc_bb3_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_stop(Ar_0, Ar_1, Ar_2))
	eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb3_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 > Ar_1 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    eval_abc_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb0_in(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_bb0_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_0(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_0(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_1(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_1(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_2(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_2(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_3(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_3(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_4(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)    eval_abc_4(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_0))
		(Comp: ?, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb2_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 <= Ar_1 ]
		(Comp: 2, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb3_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 > Ar_1 ]
		(Comp: ?, Cost: 1)    eval_abc_bb2_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_2 + 1))
		(Comp: 2, Cost: 1)    eval_abc_bb3_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_stop(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(eval_abc_start) = -2*V_1 + 2*V_2 + 2
	Pol(eval_abc_bb0_in) = -2*V_1 + 2*V_2 + 2
	Pol(eval_abc_0) = -2*V_1 + 2*V_2 + 2
	Pol(eval_abc_1) = -2*V_1 + 2*V_2 + 2
	Pol(eval_abc_2) = -2*V_1 + 2*V_2 + 2
	Pol(eval_abc_3) = -2*V_1 + 2*V_2 + 2
	Pol(eval_abc_4) = -2*V_1 + 2*V_2 + 2
	Pol(eval_abc_bb1_in) = 2*V_2 - 2*V_3 + 2
	Pol(eval_abc_bb2_in) = 2*V_2 - 2*V_3 + 1
	Pol(eval_abc_bb3_in) = 2*V_2 - 2*V_3 + 2
	Pol(eval_abc_stop) = 2*V_2 - 2*V_3 + 2
	Pol(koat_start) = -2*V_1 + 2*V_2 + 2
orients all transitions weakly and the transition
	eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb2_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 <= Ar_1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 1)                      eval_abc_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb0_in(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_bb0_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_0(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_0(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_1(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_1(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_2(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_2(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_3(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_3(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_4(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_4(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_0))
		(Comp: 2*Ar_0 + 2*Ar_1 + 2, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb2_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 <= Ar_1 ]
		(Comp: 2, Cost: 1)                      eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb3_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 > Ar_1 ]
		(Comp: ?, Cost: 1)                      eval_abc_bb2_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_2 + 1))
		(Comp: 2, Cost: 1)                      eval_abc_bb3_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_stop(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 0)                      koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 4 produces the following problem:
5:	T:
		(Comp: 1, Cost: 1)                      eval_abc_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb0_in(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_bb0_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_0(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_0(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_1(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_1(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_2(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_2(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_3(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_3(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_4(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 1)                      eval_abc_4(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_0))
		(Comp: 2*Ar_0 + 2*Ar_1 + 2, Cost: 1)    eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb2_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 <= Ar_1 ]
		(Comp: 2, Cost: 1)                      eval_abc_bb1_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb3_in(Ar_0, Ar_1, Ar_2)) [ Ar_2 > Ar_1 ]
		(Comp: 2*Ar_0 + 2*Ar_1 + 2, Cost: 1)    eval_abc_bb2_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_bb1_in(Ar_0, Ar_1, Ar_2 + 1))
		(Comp: 2, Cost: 1)                      eval_abc_bb3_in(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_stop(Ar_0, Ar_1, Ar_2))
		(Comp: 1, Cost: 0)                      koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(eval_abc_start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 4*Ar_0 + 4*Ar_1 + 15

Time: 0.663 sec (SMT: 0.642 sec)