
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f6) = 1
	Pol(f14) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transitions
	f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
	f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
	f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ 0 >= Ar_0 + 1 ]
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = V_2
	Pol(f6) = V_1
	Pol(f14) = 0
	Pol(koat_start) = V_2
orients all transitions weakly and the transition
	f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_0 >= 1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3))
		(Comp: ?, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ 0 >= Ar_0 + 1 ]
		(Comp: Ar_1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f6: -X_3 + X_4 >= 0 /\ -X_1 + X_2 >= 0


This yielded the following problem:
5:	T:
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: Ar_1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3))
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] with all transitions in problem 5, the following new transition is obtained:
	koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3)) [ 0 <= 0 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 1)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: Ar_1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ 0 >= Ar_0 + 1 ]
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 6:
	f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3))
We thus obtain the following problem:
7:	T:
		(Comp: ?, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ 0 >= Ar_0 + 1 ]
		(Comp: Ar_1, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0 - 1, Ar_1, Ar_2 - 1, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_3 >= Ar_1 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_1 >= Ar_3 + 1 /\ Ar_0 = 0 ]
		(Comp: 1, Cost: 1)       f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(0, Ar_1, Ar_2, Ar_1)) [ -Ar_2 + Ar_3 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 = 0 /\ Ar_1 = Ar_3 ]
		(Comp: 1, Cost: 1)       koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_1, Ar_1, Ar_3, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.092 sec (SMT: 0.074 sec)
