
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f16(Ar_0, 0, Fresh_2, Fresh_2, Ar_4)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= 1 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))
		(Comp: ?, Cost: 1)    f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f30(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f10(Fresh_1, Ar_1, Fresh_1, Ar_3, 0)) [ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f10(Fresh_0, 0, Fresh_0, Ar_3, 0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_0, Ar_3].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3))
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f27(Ar_0, Ar_3) -> Com_1(f30(Ar_0, Ar_3))
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3))
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 2:
	f27(Ar_0, Ar_3) -> Com_1(f30(Ar_0, Ar_3))
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3))
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 3 produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3))
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f16) = 1
	Pol(f10) = 1
	Pol(f25) = 0
	Pol(f0) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
strictly and produces the following problem:
5:	T:
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3))
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 5 to obtain the following invariants:
  For symbol f16: -X_1 >= 0
  For symbol f25: X_1 - 1 >= 0


This yielded the following problem:
6:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3))
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 - 1 >= 0 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ -Ar_0 >= 0 /\ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ -Ar_0 >= 0 /\ Ar_3 >= 1 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ] with all transitions in problem 6, the following new transition is obtained:
	koat_start(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) [ 0 <= 0 ]
We thus obtain the following problem:
7:	T:
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3))
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 - 1 >= 0 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ -Ar_0 >= 0 /\ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ -Ar_0 >= 0 /\ Ar_3 >= 1 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transition from problem 7:
	f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3))
We thus obtain the following problem:
8:	T:
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ -Ar_0 >= 0 /\ Ar_3 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ -Ar_0 >= 0 /\ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 - 1 >= 0 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ] with all transitions in problem 8, the following new transition is obtained:
	f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 /\ Ar_0 - 1 >= 0 ]
We thus obtain the following problem:
9:	T:
		(Comp: 1, Cost: 2)    f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 /\ Ar_0 - 1 >= 0 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ -Ar_0 >= 0 /\ Ar_3 >= 1 ]
		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ -Ar_0 >= 0 /\ 0 >= Ar_3 ]
		(Comp: ?, Cost: 1)    f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 - 1 >= 0 ]
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ]
		(Comp: 1, Cost: 1)    koat_start(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.042 sec (SMT: 0.032 sec)
