MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_24, Fresh_25)) [ Fresh_24 >= 1 /\ Fresh_24 >= 5 /\ Fresh_24 >= 2 /\ Fresh_24 >= 3 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_23)) [ 0 >= 4 /\ 0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_21, Fresh_22)) [ Fresh_21 >= 1 /\ Fresh_21 >= 5 /\ 0 >= Fresh_21 /\ Fresh_21 >= 3 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_20)) [ 0 >= 4 /\ 0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(3, Fresh_19))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_18)) [ 0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(3, Fresh_17)) [ 0 >= 3 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_16)) [ 0 >= 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_15)) [ Ar_1 >= 5 /\ Ar_1 >= 2 /\ Ar_1 >= 3 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_14)) [ Ar_1 >= 5 /\ Ar_1 >= 2 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_13)) [ Ar_1 >= 5 /\ 0 >= Ar_1 /\ Ar_1 >= 3 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_12)) [ Ar_1 >= 5 /\ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_11)) [ Ar_1 = 3 /\ Ar_0 = 6 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_10)) [ 3 >= Ar_1 /\ Ar_1 >= 2 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_9)) [ 0 >= 3 /\ Ar_1 = 3 /\ Ar_0 = 6 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_8)) [ 3 >= Ar_1 /\ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_7)) [ 6*Ar_1 >= 1 /\ 6*Ar_1 + 2 >= 0 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_6)) [ 6*Ar_1 >= 1 /\ 6*Ar_1 + 2 >= 0 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_5)) [ 6*Ar_1 >= 1 /\ 0 >= 6*Ar_1 + 4 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_4)) [ 6*Ar_1 >= 1 /\ 0 >= 6*Ar_1 + 4 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_3)) [ 0 >= 6*Ar_1 + 1 /\ 6*Ar_1 + 2 >= 0 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_2)) [ 0 >= 6*Ar_1 + 1 /\ 6*Ar_1 + 2 >= 0 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_1)) [ 0 >= 6*Ar_1 + 1 /\ 0 >= 6*Ar_1 + 4 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_0)) [ 0 >= 6*Ar_1 + 1 /\ 0 >= 6*Ar_1 + 4 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
		(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

Testing for reachability in the complexity graph removes the following transitions from problem 1:
	f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_23)) [ 0 >= 4 /\ 0 >= 1 ]
	f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_21, Fresh_22)) [ Fresh_21 >= 1 /\ Fresh_21 >= 5 /\ 0 >= Fresh_21 /\ Fresh_21 >= 3 ]
	f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_20)) [ 0 >= 4 /\ 0 >= 1 ]
	f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_18)) [ 0 >= 1 ]
	f0(Ar_0, Ar_1) -> Com_1(f2(3, Fresh_17)) [ 0 >= 3 ]
	f0(Ar_0, Ar_1) -> Com_1(f2(1, Fresh_16)) [ 0 >= 1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_14)) [ Ar_1 >= 5 /\ Ar_1 >= 2 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_13)) [ Ar_1 >= 5 /\ 0 >= Ar_1 /\ Ar_1 >= 3 /\ Ar_0 = 2*Ar_1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_12)) [ Ar_1 >= 5 /\ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_10)) [ 3 >= Ar_1 /\ Ar_1 >= 2 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_9)) [ 0 >= 3 /\ Ar_1 = 3 /\ Ar_0 = 6 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_8)) [ 3 >= Ar_1 /\ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 2*Ar_1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_6)) [ 6*Ar_1 >= 1 /\ 6*Ar_1 + 2 >= 0 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_5)) [ 6*Ar_1 >= 1 /\ 0 >= 6*Ar_1 + 4 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_4)) [ 6*Ar_1 >= 1 /\ 0 >= 6*Ar_1 + 4 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_3)) [ 0 >= 6*Ar_1 + 1 /\ 6*Ar_1 + 2 >= 0 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_2)) [ 0 >= 6*Ar_1 + 1 /\ 6*Ar_1 + 2 >= 0 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_1)) [ 0 >= 6*Ar_1 + 1 /\ 0 >= 6*Ar_1 + 4 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
	f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_0)) [ 0 >= 6*Ar_1 + 1 /\ 0 >= 6*Ar_1 + 4 /\ 0 >= 6*Ar_1 + 3 /\ Ar_0 = 2*Ar_1 + 1 ]
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_7)) [ 6*Ar_1 >= 1 /\ 6*Ar_1 + 2 >= 0 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_11)) [ Ar_1 = 3 /\ Ar_0 = 6 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_15)) [ Ar_1 >= 5 /\ Ar_1 >= 2 /\ Ar_1 >= 3 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(3, Fresh_19))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_24, Fresh_25)) [ Fresh_24 >= 1 /\ Fresh_24 >= 5 /\ Fresh_24 >= 2 /\ Fresh_24 >= 3 ]
		(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 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_7)) [ 6*Ar_1 >= 1 /\ 6*Ar_1 + 2 >= 0 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_11)) [ Ar_1 = 3 /\ Ar_0 = 6 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_15)) [ Ar_1 >= 5 /\ Ar_1 >= 2 /\ Ar_1 >= 3 /\ Ar_0 = 2*Ar_1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(3, Fresh_19))
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(Fresh_24, Fresh_25)) [ Fresh_24 >= 1 /\ Fresh_24 >= 5 /\ Fresh_24 >= 2 /\ Fresh_24 >= 3 ]
		(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 3 to obtain the following invariants:
  For symbol f2: X_1 - 3 >= 0


This yielded the following problem:
4:	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(f2(Fresh_24, Fresh_25)) [ Fresh_24 >= 1 /\ Fresh_24 >= 5 /\ Fresh_24 >= 2 /\ Fresh_24 >= 3 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f2(3, Fresh_19))
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_15)) [ Ar_0 - 3 >= 0 /\ Ar_1 >= 5 /\ Ar_1 >= 2 /\ Ar_1 >= 3 /\ Ar_0 = 2*Ar_1 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(Ar_1, Fresh_11)) [ Ar_0 - 3 >= 0 /\ Ar_1 = 3 /\ Ar_0 = 6 ]
		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1) -> Com_1(f2(6*Ar_1 + 4, Fresh_7)) [ Ar_0 - 3 >= 0 /\ 6*Ar_1 >= 1 /\ 6*Ar_1 + 2 >= 0 /\ 6*Ar_1 + 1 >= 0 /\ Ar_0 = 2*Ar_1 + 1 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 6.375 sec (SMT: 6.254 sec)