/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 16 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
l0(A, B) -> Com_1(l1(A, B)) :|: TRUE
l1(A, B) -> Com_1(l1(A, B - A - A + 1)) :|: B <= A && A + B >= 1

The start-symbols are:[l0_2]


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(1) Koat Proof (FINISHED)
YES(?, 3)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 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)    l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1))

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1)) with all transitions in problem 2, the following new transition is obtained:

	l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ]

We thus obtain the following problem:

3:	T:

		(Comp: 1, Cost: 2)    l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ]

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



By chaining the transition l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ] with all transitions in problem 3, the following new transition is obtained:

	l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 4*Ar_0 + 2)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 /\ Ar_0 >= Ar_1 - 2*Ar_0 + 1 /\ -Ar_0 + Ar_1 + 1 >= 1 ]

We thus obtain the following problem:

4:	T:

		(Comp: 1, Cost: 3)    l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 4*Ar_0 + 2)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 /\ Ar_0 >= Ar_1 - 2*Ar_0 + 1 /\ -Ar_0 + Ar_1 + 1 >= 1 ]

		(Comp: ?, Cost: 1)    l1(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Testing for reachability in the complexity graph removes the following transition from problem 4:

	l1(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 2*Ar_0 + 1)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 ]

We thus obtain the following problem:

5:	T:

		(Comp: 1, Cost: 3)    l0(Ar_0, Ar_1) -> Com_1(l1(Ar_0, Ar_1 - 4*Ar_0 + 2)) [ Ar_0 >= Ar_1 /\ Ar_0 + Ar_1 >= 1 /\ Ar_0 >= Ar_1 - 2*Ar_0 + 1 /\ -Ar_0 + Ar_1 + 1 >= 1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1) -> Com_1(l0(Ar_0, Ar_1)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 3



Time: 0.149 sec (SMT: 0.138 sec)


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(2)
BOUNDS(1, 1)