
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 19 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 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) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 19 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f0) = 1
	Pol(f5) = 1
	Pol(f12) = 0
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ]
strictly and produces the following problem:
3:	T:
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 19 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f5: X_1 - 1 >= 0


This yielded the following problem:
4:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] with all transitions in problem 4, the following new transitions are obtained:
	koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ]
	koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ]
We thus obtain the following problem:
5:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 5:
	f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ]
	f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ]
We thus obtain the following problem:
6:	T:
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ] with all transitions in problem 6, the following new transitions are obtained:
	koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ 19 >= Fresh_1 ]
	koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ Fresh_1 >= 20 ]
We thus obtain the following problem:
7:	T:
		(Comp: 1, Cost: 2)    koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ 19 >= Fresh_1 ]
		(Comp: 1, Cost: 2)    koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ Fresh_1 >= 20 ]
		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f5) = -V_1 + 20
and size complexities
	S("koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\\ 0 >= Fresh_0 ]", 0-0) = ?
	S("f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\\ Ar_0 >= 20 ]", 0-0) = 20
	S("f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\\ 19 >= Ar_0 ]", 0-0) = 20
	S("koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\\ Fresh_1 >= 1 /\\ Fresh_1 - 1 >= 0 /\\ Fresh_1 >= 20 ]", 0-0) = ?
	S("koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\\ Fresh_1 >= 1 /\\ Fresh_1 - 1 >= 0 /\\ 19 >= Fresh_1 ]", 0-0) = 20
orients the transitions
	f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ]
weakly and the transition
	f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ]
strictly and produces the following problem:
8:	T:
		(Comp: 1, Cost: 2)     koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ 19 >= Fresh_1 ]
		(Comp: 1, Cost: 2)     koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ Fresh_1 >= 20 ]
		(Comp: 40, Cost: 1)    f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ]
		(Comp: 1, Cost: 1)     f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ]
		(Comp: 1, Cost: 1)     koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound 46

Time: 0.055 sec (SMT: 0.050 sec)
