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Macro in the article field

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I am planning a project and I am thinking of extracting articles from a folder using macros.
Do I remember correctly that the SER item duplicate check feature does not work if I pull items from a folder?
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soft question – Error in a Wikipedia article on forbidden graph substructures

I'm sorry if that's a trivial matter. But I was hoping for clarification on what I think is a mistake in an article on wikipedia that I was reading.

To explain, given two classes of graphs $ mathcal {F} _1 $ and $ mathcal {F} _2 $ to write $ mathcal {F} _1 cong mathcal {F} _2 $ and say $ mathcal {F} _1 $ is isomorphic to $ mathcal {F} _2 $ if there is a bijection $ f $ of $ mathcal {F} _1 $ at $ mathcal {F} _2 $ so that for all $ G in mathcal {F} _1 $ we have $ f (G) cong G $ say the same $ mathcal {F} _1 preceq mathcal {F} _2 $ if and only if $ forall G in mathcal {F} _1 exists H in mathcal {F} _2: G cong H $. (This can be extended to the appropriate classes using these ideas) Now, given any class of graphics $ C $ and any pre-order $ leq $ so that if $ G in C $ and $ H leq $ G then $ H $ in C $ (that is to say. $ C $ is an ideal of $ leq $ on the class of all the graphs) we will say another class of graphs $ F $ clogs $ C $ under $ leq $ When we have $ G in C $ if and only if there is no graph $ H leq $ G which is isomorphic to a graph $ F $.

For example if every time a graph $ H $ is a minor graphic of a graph $ G $ we write that $ H leq $ G then by Wagner's theorem if $ C $ is the class of graphs then plans $ {K_5, K_ {3,3} } $ clogs $ C $ under $ leq $.

That being said, is not it true that there is a unique $ preceq $minimal obstruction (up to isomorphism) defined for any class of graphs $ C $ under any pre-order $ leq $ if and only if $ (1) $ $ C $ is an ideal of $ leq $ and $ (2) $ the order of the quotient $ leq / $ is based on the graph class not $ C $?

In this case, the whole set of single obstruction up to isomorphism would be any maximum class of inclusion $ F $ non isomorphic graphs per pair not in $ C $ so if $ G in F $ so for all $ H prec G $ we have $ H $ in C $ this is to say:

enter the description of the image here

Now, it sounds trivial but https://en.wikipedia.org/wiki/Forbidden_graph_characterization simply assumes the condition $ (2) $ is always true Which does not seem right to me, unless we only work with finished graphs. Am I wrong here? Where is wikipedia?


The proof would be essentially if $ C $ had a unique up to minimal isomorphism $ preceq $ set of obstruction and the order of the quotient $ leq / $ was not well ordered on the class of graphics not $ C $ then for any obstruction set $ F $ of $ C $ under $ leq $ it exists $ G in F implies G not in C $ so that for some $ H leq $ G we have $ H not in C $ as good as $ H not cong G $ now $ F = (F setminus {G }) cup {H } $ is a set of obstruction of $ C $ and $ F $ F $ with $ F $ F $ contradicting our hypothesis that $ C $ would have a unique $ preceq $ up to the entire obstruction of isomorphism, even if $ leq / $ is based on the graph class not $ C $ then for any obstruction set $ F $ of $ C $ we know for each graph $ G in F $ there must be a $ leq $ minimal isomorphism chart up to $ G $ leq on the class of non graphs $ C $ so that for all the graphics $ H prec G $ we have $ H $ in C $ now, if we define $ F = {G in G $ $ we see $ F & # 39; form a set of unique isomorphism obstruction since each set of obstruction $ F & # 39; $ must have $ F & # 39; preceq F & # 39; & # 39; $ and if we delete any graph of $ F & # 39; then the deleted chart would not be comparable to any of the remaining charts and therefore could not interfere $ C $ under $ leq $ which means that there is no class $ F & # 39; & # 39; prec F & # 39; $ such as $ F & # 39; & # 39; $ is a set of obstruction of $ C $ under $ leq $ So $ F & # 39; is $ preceq $ minimal. Although the fact that $ C $ must be an ideal of $ leq $ is trivial, so putting these together gives us the biconditional for $ (1) $ and $ (2) $.

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by: douglasmgaloyan
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