
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 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)    f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f3) = 1
	Pol(f0) = 1
	Pol(f4) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f3) = V_2
	Pol(f0) = V_2
	Pol(f4) = V_2
	Pol(koat_start) = V_2
orients all transitions weakly and the transition
	f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ]
strictly and produces the following problem:
4:	T:
		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1))
		(Comp: Ar_1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ]
		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ]
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f0: -X_1 >= 0 /\ X_1 >= 0
  For symbol f4: -X_2 >= 0 /\ X_1 - X_2 + 1 >= 0 /\ -X_1 - X_2 - 1 >= 0 /\ -X_1 - 1 >= 0 /\ X_1 + 1 >= 0


This yielded the following problem:
5:	T:
		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_0 - Ar_1 + 1 >= 0 /\ -Ar_0 - Ar_1 - 1 >= 0 /\ -Ar_0 - 1 >= 0 /\ Ar_0 + 1 >= 0 ]
		(Comp: Ar_1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1))
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] with all transitions in problem 5, the following new transition is obtained:
	koat_start(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) [ 0 <= 0 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 1)       koat_start(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_1 ]
		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_0 - Ar_1 + 1 >= 0 /\ -Ar_0 - Ar_1 - 1 >= 0 /\ -Ar_0 - 1 >= 0 /\ Ar_0 + 1 >= 0 ]
		(Comp: Ar_1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)       f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1))
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 6:
	f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1))
We thus obtain the following problem:
7:	T:
		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_0 - Ar_1 + 1 >= 0 /\ -Ar_0 - Ar_1 - 1 >= 0 /\ -Ar_0 - 1 >= 0 /\ Ar_0 + 1 >= 0 ]
		(Comp: Ar_1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_1 ]
		(Comp: 1, Cost: 1)       koat_start(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_1 ] with all transitions in problem 7, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_1 /\ -Ar_1 >= 0 /\ 0 >= 0 ]
We thus obtain the following problem:
8:	T:
		(Comp: 1, Cost: 2)       f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_1 /\ -Ar_1 >= 0 /\ 0 >= 0 ]
		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_0 - Ar_1 + 1 >= 0 /\ -Ar_0 - Ar_1 - 1 >= 0 /\ -Ar_0 - 1 >= 0 /\ Ar_0 + 1 >= 0 ]
		(Comp: Ar_1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 1 ]
		(Comp: 1, Cost: 1)       koat_start(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.050 sec (SMT: 0.041 sec)
