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


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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 13 ms]
(2) BOUNDS(1, 1)


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

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
evalndloopstart(A) -> Com_1(evalndloopentryin(A)) :|: TRUE
evalndloopentryin(A) -> Com_1(evalndloopbbin(0)) :|: TRUE
evalndloopbbin(A) -> Com_1(evalndloopbbin(B)) :|: 2 + A >= B && B >= A + 1 && 9 >= B
evalndloopbbin(A) -> Com_1(evalndloopreturnin(A)) :|: B >= A + 3
evalndloopbbin(A) -> Com_1(evalndloopreturnin(A)) :|: A >= B
evalndloopbbin(A) -> Com_1(evalndloopreturnin(A)) :|: B >= 10
evalndloopreturnin(A) -> Com_1(evalndloopstop(A)) :|: TRUE

The start-symbols are:[evalndloopstart_1]


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

(1) Koat Proof (FINISHED)
YES(?, 19)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    evalndloopstart(Ar_0) -> Com_1(evalndloopentryin(Ar_0))

		(Comp: ?, Cost: 1)    evalndloopentryin(Ar_0) -> Com_1(evalndloopbbin(0))

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopbbin(Fresh_0)) [ Ar_0 + 2 >= Fresh_0 /\ Fresh_0 >= Ar_0 + 1 /\ 9 >= Fresh_0 ]

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= Ar_0 + 3 ]

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ Ar_0 >= B ]

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= 10 ]

		(Comp: ?, Cost: 1)    evalndloopreturnin(Ar_0) -> Com_1(evalndloopstop(Ar_0))

		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(evalndloopstart(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)    evalndloopstart(Ar_0) -> Com_1(evalndloopentryin(Ar_0))

		(Comp: 1, Cost: 1)    evalndloopentryin(Ar_0) -> Com_1(evalndloopbbin(0))

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopbbin(Fresh_0)) [ Ar_0 + 2 >= Fresh_0 /\ Fresh_0 >= Ar_0 + 1 /\ 9 >= Fresh_0 ]

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= Ar_0 + 3 ]

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ Ar_0 >= B ]

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= 10 ]

		(Comp: ?, Cost: 1)    evalndloopreturnin(Ar_0) -> Com_1(evalndloopstop(Ar_0))

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

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalndloopstart) = 2

	Pol(evalndloopentryin) = 2

	Pol(evalndloopbbin) = 2

	Pol(evalndloopreturnin) = 1

	Pol(evalndloopstop) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	evalndloopreturnin(Ar_0) -> Com_1(evalndloopstop(Ar_0))

	evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= 10 ]

	evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= Ar_0 + 3 ]

	evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ Ar_0 >= B ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    evalndloopstart(Ar_0) -> Com_1(evalndloopentryin(Ar_0))

		(Comp: 1, Cost: 1)    evalndloopentryin(Ar_0) -> Com_1(evalndloopbbin(0))

		(Comp: ?, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopbbin(Fresh_0)) [ Ar_0 + 2 >= Fresh_0 /\ Fresh_0 >= Ar_0 + 1 /\ 9 >= Fresh_0 ]

		(Comp: 2, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= Ar_0 + 3 ]

		(Comp: 2, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ Ar_0 >= B ]

		(Comp: 2, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= 10 ]

		(Comp: 2, Cost: 1)    evalndloopreturnin(Ar_0) -> Com_1(evalndloopstop(Ar_0))

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

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(evalndloopstart) = 9

	Pol(evalndloopentryin) = 9

	Pol(evalndloopbbin) = -V_1 + 9

	Pol(evalndloopreturnin) = -V_1

	Pol(evalndloopstop) = -V_1

	Pol(koat_start) = 9

orients all transitions weakly and the transition

	evalndloopbbin(Ar_0) -> Com_1(evalndloopbbin(Fresh_0)) [ Ar_0 + 2 >= Fresh_0 /\ Fresh_0 >= Ar_0 + 1 /\ 9 >= Fresh_0 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)    evalndloopstart(Ar_0) -> Com_1(evalndloopentryin(Ar_0))

		(Comp: 1, Cost: 1)    evalndloopentryin(Ar_0) -> Com_1(evalndloopbbin(0))

		(Comp: 9, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopbbin(Fresh_0)) [ Ar_0 + 2 >= Fresh_0 /\ Fresh_0 >= Ar_0 + 1 /\ 9 >= Fresh_0 ]

		(Comp: 2, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= Ar_0 + 3 ]

		(Comp: 2, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ Ar_0 >= B ]

		(Comp: 2, Cost: 1)    evalndloopbbin(Ar_0) -> Com_1(evalndloopreturnin(Ar_0)) [ B >= 10 ]

		(Comp: 2, Cost: 1)    evalndloopreturnin(Ar_0) -> Com_1(evalndloopstop(Ar_0))

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

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 19



Time: 0.058 sec (SMT: 0.051 sec)


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

(2)
BOUNDS(1, 1)