/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files


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WORST_CASE(?, O(1))
proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 8 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A) -> Com_1(f3(0)) :|: TRUE
f3(A) -> Com_1(f3(A + 1)) :|: 9 >= A
f3(A) -> Com_1(f11(A)) :|: A >= 10 && 0 >= B + 1
f3(A) -> Com_1(f11(A)) :|: A >= 10

The start-symbols are:[f0_1]


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(1) Koat Proof (FINISHED)
YES(?, 13)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f3(0))

		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f3(Ar_0 + 1)) [ 9 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 /\ 0 >= B + 1 ]

		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 ]

		(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(f3(0))

		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f3(Ar_0 + 1)) [ 9 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 /\ 0 >= B + 1 ]

		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 ]

		(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(f3) = 1

	Pol(f11) = 0

	Pol(koat_start) = 1

orients all transitions weakly and the transitions

	f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 /\ 0 >= B + 1 ]

	f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f3(0))

		(Comp: ?, Cost: 1)    f3(Ar_0) -> Com_1(f3(Ar_0 + 1)) [ 9 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 /\ 0 >= B + 1 ]

		(Comp: 1, Cost: 1)    f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 ]

		(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) = 10

	Pol(f3) = -V_1 + 10

	Pol(f11) = -V_1

	Pol(koat_start) = 10

orients all transitions weakly and the transition

	f3(Ar_0) -> Com_1(f3(Ar_0 + 1)) [ 9 >= Ar_0 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)     f0(Ar_0) -> Com_1(f3(0))

		(Comp: 10, Cost: 1)    f3(Ar_0) -> Com_1(f3(Ar_0 + 1)) [ 9 >= Ar_0 ]

		(Comp: 1, Cost: 1)     f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 /\ 0 >= B + 1 ]

		(Comp: 1, Cost: 1)     f3(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 10 ]

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

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 13



Time: 0.055 sec (SMT: 0.053 sec)


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(2)
BOUNDS(1, 1)