/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


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WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 203 ms]
(2) BOUNDS(1, n^1)
(3) Loat Proof [FINISHED, 1598 ms]
(4) BOUNDS(n^1, INF)


----------------------------------------

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D) -> Com_1(f4(0, B, C, D)) :|: TRUE
f4(A, B, C, D) -> Com_1(f8(A + 1, B, 0, D)) :|: B >= A + 1
f8(A, B, C, D) -> Com_1(f16(A, B, C, 0)) :|: B >= A + 1
f8(A, B, C, D) -> Com_1(f16(A, B, C, D)) :|: A >= B
f8(A, B, C, D) -> Com_1(f8(A + 1, B, C + 1, E)) :|: B >= A + 1 && 0 >= E + 1
f8(A, B, C, D) -> Com_1(f8(A + 1, B, C + 1, E)) :|: B >= A + 1 && E >= 1
f16(A, B, C, D) -> Com_1(f4(A, B, C, D)) :|: 0 >= C
f16(A, B, C, D) -> Com_1(f4(A - 1, B, C, D)) :|: C >= 1
f4(A, B, C, D) -> Com_1(f20(A, B, C, D)) :|: A >= B

The start-symbols are:[f0_4]


----------------------------------------

(1) Koat Proof (FINISHED)
YES(?, 17*Ar_1 + 2)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(0, Ar_1, Ar_2, Ar_3))

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f8(Ar_0 + 1, Ar_1, 0, Ar_3)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, 0)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1, Fresh_1)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1, Fresh_0)) [ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f20(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 ]

		(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



Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_0, Ar_1, Ar_2].

We thus obtain the following problem:

2:	T:

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, 0)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, Ar_1, Ar_2))

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 2 produces the following problem:

3:	T:

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, 0)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, Ar_1, Ar_2))

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(koat_start) = 1

	Pol(f0) = 1

	Pol(f4) = 1

	Pol(f20) = 0

	Pol(f16) = 1

	Pol(f8) = 1

orients all transitions weakly and the transition

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: ?, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, 0)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, Ar_1, Ar_2))

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(koat_start) = V_2

	Pol(f0) = V_2 - 1

	Pol(f4) = -V_1 + V_2 - 1

	Pol(f20) = -V_1 + V_2 - 1

	Pol(f16) = -V_1 + V_2

	Pol(f8) = -V_1 + V_2

orients all transitions weakly and the transitions

	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

		(Comp: 1, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)       f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_2 >= 1 ]

		(Comp: ?, Cost: 1)       f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]

		(Comp: Ar_1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

		(Comp: Ar_1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

		(Comp: ?, Cost: 1)       f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ]

		(Comp: ?, Cost: 1)       f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, 0)) [ Ar_1 >= Ar_0 + 1 ]

		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, Ar_1, Ar_2))

	start location:	koat_start

	leaf cost:	0



Applied AI with 'oct' on problem 5 to obtain the following invariants:

  For symbol f16: X_2 - X_3 - 1 >= 0 /\ X_1 - X_3 - 1 >= 0 /\ X_3 >= 0 /\ X_2 + X_3 - 1 >= 0 /\ X_1 + X_3 - 1 >= 0 /\ X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ -X_1 + X_2 >= 0 /\ X_1 - 1 >= 0

  For symbol f4: X_1 >= 0

  For symbol f8: X_2 - X_3 - 1 >= 0 /\ X_1 - X_3 - 1 >= 0 /\ X_3 >= 0 /\ X_2 + X_3 - 1 >= 0 /\ X_1 + X_3 - 1 >= 0 /\ X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ -X_1 + X_2 >= 0 /\ X_1 - 1 >= 0





This yielded the following problem:

6:	T:

		(Comp: 1, Cost: 1)       f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, Ar_1, Ar_2))

		(Comp: ?, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)       f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)       f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= Ar_1 ]

		(Comp: Ar_1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

		(Comp: Ar_1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

		(Comp: ?, Cost: 1)       f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ 0 >= Ar_2 ]

		(Comp: ?, Cost: 1)       f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_2 >= 1 ]

		(Comp: 1, Cost: 1)       f4(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 0 /\ Ar_0 >= Ar_1 ]

		(Comp: 1, Cost: 0)       koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f0) = 3*V_2

	Pol(f4) = -3*V_1 + 3*V_2

	Pol(f8) = -3*V_1 + 3*V_2 + 3*V_3 + 2

	Pol(f16) = -3*V_1 + 3*V_2 + 3*V_3 + 1

	Pol(f20) = -3*V_1 + 3*V_2

	Pol(koat_start) = 3*V_2

orients all transitions weakly and the transitions

	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= Ar_1 ]

	f4(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

	f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ 0 >= Ar_2 ]

	f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_2 >= 1 ]

strictly and produces the following problem:

7:	T:

		(Comp: 1, Cost: 1)         f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(0, Ar_1, Ar_2))

		(Comp: 3*Ar_1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, 0)) [ Ar_0 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

		(Comp: 3*Ar_1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_1 >= Ar_0 + 1 ]

		(Comp: 3*Ar_1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f16(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= Ar_1 ]

		(Comp: Ar_1, Cost: 1)      f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_1 + 1 ]

		(Comp: Ar_1, Cost: 1)      f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 1, Ar_1, Ar_2 + 1)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_1 >= Ar_0 + 1 /\ Fresh_0 >= 1 ]

		(Comp: 3*Ar_1, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ 0 >= Ar_2 ]

		(Comp: 3*Ar_1, Cost: 1)    f16(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0 - 1, Ar_1, Ar_2)) [ Ar_1 - Ar_2 - 1 >= 0 /\ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ Ar_1 - 1 >= 0 /\ Ar_0 + Ar_1 - 2 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_2 >= 1 ]

		(Comp: 1, Cost: 1)         f4(Ar_0, Ar_1, Ar_2) -> Com_1(f20(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 0 /\ Ar_0 >= Ar_1 ]

		(Comp: 1, Cost: 0)         koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 17*Ar_1 + 2



Time: 0.292 sec (SMT: 0.243 sec)


----------------------------------------

(2)
BOUNDS(1, n^1)

----------------------------------------

(3) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f8 : A'=1+A, C'=0, [ B>=1+A ], cost: 1

      8: f4 -> f20 : [ A>=B ], cost: 1

      2: f8 -> f16 : D'=0, [ B>=1+A ], cost: 1

      3: f8 -> f16 : [ A>=B ], cost: 1

      4: f8 -> f8 : A'=1+A, C'=1+C, D'=free, [ B>=1+A && 0>=1+free ], cost: 1

      5: f8 -> f8 : A'=1+A, C'=1+C, D'=free_1, [ B>=1+A && free_1>=1 ], cost: 1

      6: f16 -> f4 : [ 0>=C ], cost: 1

      7: f16 -> f4 : A'=-1+A, [ C>=1 ], cost: 1



Checking for constant complexity:

   The following rule is satisfiable with cost >= 1, yielding constant complexity:

      0: f0 -> f4 : A'=0, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f8 : A'=1+A, C'=0, [ B>=1+A ], cost: 1

      2: f8 -> f16 : D'=0, [ B>=1+A ], cost: 1

      3: f8 -> f16 : [ A>=B ], cost: 1

      4: f8 -> f8 : A'=1+A, C'=1+C, D'=free, [ B>=1+A && 0>=1+free ], cost: 1

      5: f8 -> f8 : A'=1+A, C'=1+C, D'=free_1, [ B>=1+A && free_1>=1 ], cost: 1

      6: f16 -> f4 : [ 0>=C ], cost: 1

      7: f16 -> f4 : A'=-1+A, [ C>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 2.

   Accelerating the following rules:

      4: f8 -> f8 : A'=1+A, C'=1+C, D'=free, [ B>=1+A && 0>=1+free ], cost: 1

      5: f8 -> f8 : A'=1+A, C'=1+C, D'=free_1, [ B>=1+A && free_1>=1 ], cost: 1



   Accelerated rule 4 with metering function -A+B, yielding the new rule 9.

   Accelerated rule 5 with metering function -A+B, yielding the new rule 10.

   Removing the simple loops: 4 5.



Accelerated all simple loops using metering functions (where possible):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f8 : A'=1+A, C'=0, [ B>=1+A ], cost: 1

      2: f8 -> f16 : D'=0, [ B>=1+A ], cost: 1

      3: f8 -> f16 : [ A>=B ], cost: 1

      9: f8 -> f8 : A'=B, C'=C-A+B, D'=free, [ B>=1+A && 0>=1+free ], cost: -A+B

     10: f8 -> f8 : A'=B, C'=C-A+B, D'=free_1, [ B>=1+A && free_1>=1 ], cost: -A+B

      6: f16 -> f4 : [ 0>=C ], cost: 1

      7: f16 -> f4 : A'=-1+A, [ C>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f8 : A'=1+A, C'=0, [ B>=1+A ], cost: 1

     11: f4 -> f8 : A'=B, C'=-1-A+B, D'=free, [ B>=2+A && 0>=1+free ], cost: -A+B

     12: f4 -> f8 : A'=B, C'=-1-A+B, D'=free_1, [ B>=2+A && free_1>=1 ], cost: -A+B

      2: f8 -> f16 : D'=0, [ B>=1+A ], cost: 1

      3: f8 -> f16 : [ A>=B ], cost: 1

      6: f16 -> f4 : [ 0>=C ], cost: 1

      7: f16 -> f4 : A'=-1+A, [ C>=1 ], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

     13: f4 -> f16 : A'=1+A, C'=0, D'=0, [ B>=2+A ], cost: 2

     14: f4 -> f16 : A'=1+A, C'=0, [ B>=1+A && 1+A>=B ], cost: 2

     15: f4 -> f16 : A'=B, C'=-1-A+B, D'=free, [ B>=2+A && 0>=1+free ], cost: 1-A+B

     16: f4 -> f16 : A'=B, C'=-1-A+B, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 1-A+B

     17: f4 -> [6] : [ B>=2+A && 0>=1+free ], cost: -A+B

     18: f4 -> [6] : [ B>=2+A && free_1>=1 ], cost: -A+B

      6: f16 -> f4 : [ 0>=C ], cost: 1

      7: f16 -> f4 : A'=-1+A, [ C>=1 ], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

     17: f4 -> [6] : [ B>=2+A && 0>=1+free ], cost: -A+B

     18: f4 -> [6] : [ B>=2+A && free_1>=1 ], cost: -A+B

     19: f4 -> f4 : A'=1+A, C'=0, D'=0, [ B>=2+A ], cost: 3

     20: f4 -> f4 : A'=1+A, C'=0, [ B>=1+A && 1+A>=B ], cost: 3

     21: f4 -> f4 : A'=-1+B, C'=-1-A+B, D'=free, [ B>=2+A && 0>=1+free ], cost: 2-A+B

     22: f4 -> f4 : A'=-1+B, C'=-1-A+B, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 2-A+B

     23: f4 -> [7] : [ B>=2+A && 0>=1+free ], cost: 1-A+B

     24: f4 -> [7] : [ B>=2+A && free_1>=1 ], cost: 1-A+B



Accelerating simple loops of location 1.

   Simplified some of the simple loops (and removed duplicate rules).

   Accelerating the following rules:

     19: f4 -> f4 : A'=1+A, C'=0, D'=0, [ B>=2+A ], cost: 3

     20: f4 -> f4 : A'=1+A, C'=0, [ 1+A-B==0 ], cost: 3

     21: f4 -> f4 : A'=-1+B, C'=-1-A+B, D'=free, [ B>=2+A && 0>=1+free ], cost: 2-A+B

     22: f4 -> f4 : A'=-1+B, C'=-1-A+B, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 2-A+B



   Accelerated rule 19 with metering function -1-A+B, yielding the new rule 25.

   Accelerated rule 20 with metering function -A+B, yielding the new rule 26.

   Found no metering function for rule 21.

   Found no metering function for rule 22.

   Removing the simple loops: 19 20.



Accelerated all simple loops using metering functions (where possible):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

     17: f4 -> [6] : [ B>=2+A && 0>=1+free ], cost: -A+B

     18: f4 -> [6] : [ B>=2+A && free_1>=1 ], cost: -A+B

     21: f4 -> f4 : A'=-1+B, C'=-1-A+B, D'=free, [ B>=2+A && 0>=1+free ], cost: 2-A+B

     22: f4 -> f4 : A'=-1+B, C'=-1-A+B, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 2-A+B

     23: f4 -> [7] : [ B>=2+A && 0>=1+free ], cost: 1-A+B

     24: f4 -> [7] : [ B>=2+A && free_1>=1 ], cost: 1-A+B

     25: f4 -> f4 : A'=-1+B, C'=0, D'=0, [ B>=2+A ], cost: -3-3*A+3*B

     26: f4 -> f4 : A'=B, C'=0, [ 1+A-B==0 ], cost: -3*A+3*B



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

     27: f0 -> f4 : A'=-1+B, C'=-1+B, D'=free, [ B>=2 && 0>=1+free ], cost: 3+B

     28: f0 -> f4 : A'=-1+B, C'=-1+B, D'=free_1, [ B>=2 && free_1>=1 ], cost: 3+B

     29: f0 -> f4 : A'=-1+B, C'=0, D'=0, [ B>=2 ], cost: -2+3*B

     30: f0 -> f4 : A'=B, C'=0, [ 1-B==0 ], cost: 1+3*B

     17: f4 -> [6] : [ B>=2+A && 0>=1+free ], cost: -A+B

     18: f4 -> [6] : [ B>=2+A && free_1>=1 ], cost: -A+B

     23: f4 -> [7] : [ B>=2+A && 0>=1+free ], cost: 1-A+B

     24: f4 -> [7] : [ B>=2+A && free_1>=1 ], cost: 1-A+B



Eliminated locations (on tree-shaped paths):

   Start location: f0

     31: f0 -> [6] : A'=0, [ B>=2 && 0>=1+free ], cost: 1+B

     32: f0 -> [6] : A'=0, [ B>=2 && free_1>=1 ], cost: 1+B

     33: f0 -> [7] : A'=0, [ B>=2 && 0>=1+free ], cost: 2+B

     34: f0 -> [7] : A'=0, [ B>=2 && free_1>=1 ], cost: 2+B

     35: f0 -> [9] : [ B>=2 && 0>=1+free ], cost: 3+B

     36: f0 -> [9] : [ B>=2 && free_1>=1 ], cost: 3+B

     37: f0 -> [9] : [ B>=2 ], cost: -2+3*B

     38: f0 -> [9] : [ 1-B==0 ], cost: 1+3*B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

     35: f0 -> [9] : [ B>=2 && 0>=1+free ], cost: 3+B

     36: f0 -> [9] : [ B>=2 && free_1>=1 ], cost: 3+B

     37: f0 -> [9] : [ B>=2 ], cost: -2+3*B

     38: f0 -> [9] : [ 1-B==0 ], cost: 1+3*B



Computing asymptotic complexity for rule 35

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -1+B (+/+!), 3+B (+), -free (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {free==-n,B==n}

      resulting limit problem:

      [solved]



   Solution:

      free / -n

      B / n

   Resulting cost 3+n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Poly(n^1)

   Cpx degree:  1

   Solved cost: 3+n

   Rule cost:   3+B

   Rule guard:  [ B>=2 && 0>=1+free ]



WORST_CASE(Omega(n^1),?)


----------------------------------------

(4)
BOUNDS(n^1, INF)