
Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f28(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14)) [ 0 >= Ar_0 ]
		(Comp: ?, Cost: 1)    f40(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f40(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14))
		(Comp: ?, Cost: 1)    f42(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f45(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14))
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f40(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f40(Ar_0, Ar_1, 0, Fresh_5, Fresh_5, Fresh_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14)) [ 0 >= Ar_0 /\ Ar_1 + 999 >= Fresh_5 ]
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f40(1, Ar_1, 0, Fresh_4, Fresh_4, Fresh_4, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14)) [ 0 >= Ar_0 /\ Fresh_4 >= Ar_1 + 1000 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f28(0, Fresh_3, Ar_2, Ar_3, Ar_4, Ar_5, 0, Fresh_3, Fresh_3, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f20(1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Fresh_2, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14)) [ 0 >= Fresh_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f20(1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, 0, 1, Fresh_1, Fresh_1, Fresh_1, Fresh_1)) [ Fresh_1 >= 1 /\ Q >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f40(1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, 0, 1, Fresh_0, Fresh_0, Fresh_0, Fresh_0)) [ 0 >= Fresh_0 /\ Q >= 1 ]
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14)) [ 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_1].
We thus obtain the following problem:
2:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Fresh_0 /\ Q >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ Fresh_1 >= 1 /\ Q >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ 0 >= Fresh_2 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f28(0, Fresh_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Ar_0 /\ Fresh_4 >= Ar_1 + 1000 ]
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ 0 >= Ar_0 /\ Ar_1 + 999 >= Fresh_5 ]
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f42(Ar_0, Ar_1) -> Com_1(f45(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f40(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f28(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f28(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ Ar_0 >= 1 ]
	f42(Ar_0, Ar_1) -> Com_1(f45(Ar_0, Ar_1))
	f20(Ar_0, Ar_1) -> Com_1(f28(Ar_0, Ar_1)) [ 0 >= Ar_0 ]
We thus obtain the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ 0 >= Ar_0 /\ Ar_1 + 999 >= Fresh_5 ]
		(Comp: ?, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Ar_0 /\ Fresh_4 >= Ar_1 + 1000 ]
		(Comp: ?, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f28(0, Fresh_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f40(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1))
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ 0 >= Fresh_2 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ Fresh_1 >= 1 /\ Q >= 1 ]
		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Fresh_0 /\ Q >= 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

Repeatedly propagating knowledge in problem 3 produces the following problem:
4:	T:
		(Comp: 2, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ 0 >= Ar_0 /\ Ar_1 + 999 >= Fresh_5 ]
		(Comp: 2, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Ar_0 /\ Fresh_4 >= Ar_1 + 1000 ]
		(Comp: 2, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f28(0, Fresh_3)) [ Ar_0 >= 1 ]
		(Comp: ?, Cost: 1)    f40(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1))
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ 0 >= Fresh_2 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ Fresh_1 >= 1 /\ Q >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Fresh_0 /\ Q >= 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

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f20: -X_1 + 1 >= 0 /\ X_1 - 1 >= 0
  For symbol f28: -X_1 >= 0 /\ X_1 >= 0
  For symbol f40: -X_1 + 1 >= 0 /\ X_1 >= 0


This yielded the following problem:
5:	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(f40(1, Ar_1)) [ 0 >= Fresh_0 /\ Q >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ Fresh_1 >= 1 /\ Q >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ 0 >= Fresh_2 ]
		(Comp: ?, Cost: 1)    f40(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ -Ar_0 + 1 >= 0 /\ Ar_0 >= 0 ]
		(Comp: 2, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f28(0, Fresh_3)) [ -Ar_0 + 1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= 1 ]
		(Comp: 2, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_0 /\ Fresh_4 >= Ar_1 + 1000 ]
		(Comp: 2, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_0 /\ Ar_1 + 999 >= Fresh_5 ]
	start location:	koat_start
	leaf cost:	0

By chaining the transition f0(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Fresh_0 /\ Q >= 1 ] with all transitions in problem 5, the following new transition is obtained:
	f0(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Fresh_0 /\ Q >= 1 /\ 0 >= 0 /\ 1 >= 0 ]
We thus obtain the following problem:
6:	T:
		(Comp: 1, Cost: 2)    f0(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ 0 >= Fresh_0 /\ Q >= 1 /\ 0 >= 0 /\ 1 >= 0 ]
		(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(f20(1, Ar_1)) [ Fresh_1 >= 1 /\ Q >= 1 ]
		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1) -> Com_1(f20(1, Ar_1)) [ 0 >= Fresh_2 ]
		(Comp: ?, Cost: 1)    f40(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ -Ar_0 + 1 >= 0 /\ Ar_0 >= 0 ]
		(Comp: 2, Cost: 1)    f20(Ar_0, Ar_1) -> Com_1(f28(0, Fresh_3)) [ -Ar_0 + 1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= 1 ]
		(Comp: 2, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_0 /\ Fresh_4 >= Ar_1 + 1000 ]
		(Comp: 2, Cost: 1)    f28(Ar_0, Ar_1) -> Com_1(f40(Ar_0, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_0 /\ Ar_1 + 999 >= Fresh_5 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 0.040 sec (SMT: 0.031 sec)
