/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 134 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval(A) -> Com_1(eval(B)) :|: A >= 0 && B + 2 * A >= 10 && 10 >= 2 * A + B
start(A) -> Com_1(eval(A)) :|: TRUE

The start-symbols are:[start_1]


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(1) Koat Proof (FINISHED)
YES(?, 5)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    start(ar_0) -> Com_1(eval(ar_0))

		(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(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)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

		(Comp: 1, Cost: 1)    start(ar_0) -> Com_1(eval(ar_0))

		(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition start(ar_0) -> Com_1(eval(ar_0)) with all transitions in problem 2, the following new transition is obtained:

	start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

We thus obtain the following problem:

3:	T:

		(Comp: 1, Cost: 2)    start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition start(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ] with all transitions in problem 3, the following new transition is obtained:

	start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ]

We thus obtain the following problem:

4:	T:

		(Comp: 1, Cost: 3)    start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ]

		(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition start(ar_0) -> Com_1(eval(4*ar_0 - 10)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 ] with all transitions in problem 4, the following new transition is obtained:

	start(ar_0) -> Com_1(eval(-8*ar_0 + 30)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 ]

We thus obtain the following problem:

5:	T:

		(Comp: 1, Cost: 4)    start(ar_0) -> Com_1(eval(-8*ar_0 + 30)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 ]

		(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition start(ar_0) -> Com_1(eval(-8*ar_0 + 30)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 ] with all transitions in problem 5, the following new transition is obtained:

	start(ar_0) -> Com_1(eval(16*ar_0 - 50)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 /\ -8*ar_0 + 30 >= 0 ]

We thus obtain the following problem:

6:	T:

		(Comp: 1, Cost: 5)    start(ar_0) -> Com_1(eval(16*ar_0 - 50)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 /\ -8*ar_0 + 30 >= 0 ]

		(Comp: ?, Cost: 1)    eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Testing for reachability in the complexity graph removes the following transition from problem 6:

	eval(ar_0) -> Com_1(eval(-2*ar_0 + 10)) [ ar_0 >= 0 ]

We thus obtain the following problem:

7:	T:

		(Comp: 1, Cost: 5)    start(ar_0) -> Com_1(eval(16*ar_0 - 50)) [ ar_0 >= 0 /\ -2*ar_0 + 10 >= 0 /\ 4*ar_0 - 10 >= 0 /\ -8*ar_0 + 30 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0) -> Com_1(start(ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 5



Time: 0.168 sec (SMT: 0.161 sec)


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(2)
BOUNDS(1, 1)