/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(?, O(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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(1, 1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 29 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f4(A) -> Com_1(f5(A)) :|: 0 >= B + 1
f4(A) -> Com_1(f5(A)) :|: TRUE
f0(A) -> Com_1(f4(0)) :|: TRUE
f5(A) -> Com_1(f11(A)) :|: A >= 3
f4(A) -> Com_1(f11(A)) :|: TRUE
f5(A) -> Com_1(f4(A + 1)) :|: 2 >= A
f11(A) -> Com_1(f14(A)) :|: 1 >= A
f11(A) -> Com_1(f14(A)) :|: A >= 2

The start-symbols are:[f0_1]


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(1) Koat Proof (FINISHED)
YES(?, 20)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0)) [ 0 >= B + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0))

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

		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 3 ]

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f11(Ar_0))

		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 2 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ 1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ Ar_0 >= 2 ]

		(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: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0)) [ 0 >= B + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0))

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

		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 3 ]

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f11(Ar_0))

		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 2 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ 1 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ Ar_0 >= 2 ]

		(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(f4) = 2

	Pol(f5) = 2

	Pol(f0) = 2

	Pol(f11) = 1

	Pol(f14) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	f5(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 3 ]

	f4(Ar_0) -> Com_1(f11(Ar_0))

	f11(Ar_0) -> Com_1(f14(Ar_0)) [ 1 >= Ar_0 ]

	f11(Ar_0) -> Com_1(f14(Ar_0)) [ Ar_0 >= 2 ]

strictly and produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0)) [ 0 >= B + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0))

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

		(Comp: 2, Cost: 1)    f5(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 3 ]

		(Comp: 2, Cost: 1)    f4(Ar_0) -> Com_1(f11(Ar_0))

		(Comp: ?, Cost: 1)    f5(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 2 >= Ar_0 ]

		(Comp: 2, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ 1 >= Ar_0 ]

		(Comp: 2, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ Ar_0 >= 2 ]

		(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(f4) = -V_1 + 3

	Pol(f5) = -V_1 + 3

	Pol(f0) = 3

	Pol(f11) = -V_1

	Pol(f14) = -V_1

	Pol(koat_start) = 3

orients all transitions weakly and the transition

	f5(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 2 >= Ar_0 ]

strictly and produces the following problem:

4:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0)) [ 0 >= B + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0))

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

		(Comp: 2, Cost: 1)    f5(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 3 ]

		(Comp: 2, Cost: 1)    f4(Ar_0) -> Com_1(f11(Ar_0))

		(Comp: 3, Cost: 1)    f5(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 2 >= Ar_0 ]

		(Comp: 2, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ 1 >= Ar_0 ]

		(Comp: 2, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ Ar_0 >= 2 ]

		(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 4 produces the following problem:

5:	T:

		(Comp: 4, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0)) [ 0 >= B + 1 ]

		(Comp: 4, Cost: 1)    f4(Ar_0) -> Com_1(f5(Ar_0))

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

		(Comp: 2, Cost: 1)    f5(Ar_0) -> Com_1(f11(Ar_0)) [ Ar_0 >= 3 ]

		(Comp: 2, Cost: 1)    f4(Ar_0) -> Com_1(f11(Ar_0))

		(Comp: 3, Cost: 1)    f5(Ar_0) -> Com_1(f4(Ar_0 + 1)) [ 2 >= Ar_0 ]

		(Comp: 2, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ 1 >= Ar_0 ]

		(Comp: 2, Cost: 1)    f11(Ar_0) -> Com_1(f14(Ar_0)) [ Ar_0 >= 2 ]

		(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 20



Time: 0.043 sec (SMT: 0.039 sec)


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(2)
BOUNDS(1, 1)