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[Merged by Bors] - feat(Combinatorics/Graph): OrderBot instance for Graph
#37610
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init
Jun2M 378bfd6
isClosedSubgraph_bot_iff
Jun2M cad5d40
disjoint
Jun2M 5c1189e
Merge branch 'master' into GraphOrderBot
Jun2M bddb032
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Jun2M c298fd6
Update Mathlib/Combinatorics/Graph/Subgraph.lean
Jun2M 0b0786d
rm vertexSet_disjoint_of_disjoint
Jun2M 59d6981
Merge branch 'master' into GraphOrderBot
Jun2M 6c7c3b2
Merge branch 'master' into GraphOrderBot
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Sorry, I do not buy this still. For me,
G = ⊥is simpler and more useful thanV(G) = ∅: it only involves equality, and can be substituted in. I would therefore suggest you build the API aroundG = ⊥rather thanV(G) = ∅. Right now, it looks like a confusing mix of bothThere was a problem hiding this comment.
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I guess the question here is: should the negation of SNF(simp normal form) also be SNF?
In my view, answer is 'not necessarily'. "
s.Nonemptyis a great example: just becauses = ∅is SNF, you wouldn't argues ≠ ∅should be the SNF, nots.Nonempty, would you?I agree that "
G = ⊥is simpler and more useful thanV(G) = ∅". But I would also argue thatV(G).Nonemptyis more useful thanG ≠ ⊥; In almost all cases I can think of, the next step from showing a graph is nonempty is to take a vertex from the said graph. In my mind, this situation is not so different fromSet.Nonemptyone.In this perspective, there is no "mix", all equivalent forms of
G = ⊥are simplified toG = ⊥and all equivalent forms ofV(G).Nonemptyare simplified toV(G).Nonempty. The fact that those two SNF are not negation of each other is not a problem.