/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 35 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D, E) -> Com_1(f7(F, F, 10, 0, E)) :|: TRUE
f7(A, B, C, D, E) -> Com_1(f7(A, B, C, D + 1, F)) :|: C >= D + 1
f7(A, B, C, D, E) -> Com_1(f19(A, B, C, D, E)) :|: D >= C

The start-symbols are:[f0_5]


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(1) Koat Proof (FINISHED)
YES(?, 12)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_1, Fresh_1, 10, 0, Ar_4))

		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Fresh_0)) [ Ar_2 >= Ar_3 + 1 ]

		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f19(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_2 ]

		(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



Repeatedly propagating knowledge in problem 1 produces the following problem:

2:	T:

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_1, Fresh_1, 10, 0, Ar_4))

		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Fresh_0)) [ Ar_2 >= Ar_3 + 1 ]

		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f19(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_2 ]

		(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



A polynomial rank function with

	Pol(f0) = 1

	Pol(f7) = 1

	Pol(f19) = 0

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f19(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_2 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_1, Fresh_1, 10, 0, Ar_4))

		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Fresh_0)) [ Ar_2 >= Ar_3 + 1 ]

		(Comp: 1, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f19(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_2 ]

		(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



A polynomial rank function with

	Pol(f0) = 10

	Pol(f7) = V_3 - V_4

	Pol(f19) = V_3 - V_4

	Pol(koat_start) = 10

orients all transitions weakly and the transition

	f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Fresh_0)) [ Ar_2 >= Ar_3 + 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)     f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_1, Fresh_1, 10, 0, Ar_4))

		(Comp: 10, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Fresh_0)) [ Ar_2 >= Ar_3 + 1 ]

		(Comp: 1, Cost: 1)     f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f19(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_2 ]

		(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



Complexity upper bound 12



Time: 0.035 sec (SMT: 0.032 sec)


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(2)
BOUNDS(1, 1)