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I'd like to prove a proposition true over all valid Directed Acausal Graphs. I think I can do that by starting with a graph with one node and adding either a new node and connection, or a new valid connection. This seems like it would require it's own proof, though - can someone corroborate my intuition, or else show that the proof is trivial?

linkhyrule5
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  • What is your claim? – agha Jun 22 '14 at 15:03
  • Does it matter? It's rather complicated to explain, but it's a function that takes a DAG and returns a boolean, so it's valid. – linkhyrule5 Jun 22 '14 at 23:56
  • Of course it matters ! Being able to use the inductive hypothesis really depends on what you want to prove. If your inductive hypothesis states something like "if $D$ has property $X$, then it has property $Y$", then to use induction you have to be able to remove a vertex from your graph, and show that the graph has property $X$. If it does, then you can use induction to show it has property $Y$. But sometimes you can't guarantee property $X$ is there after removing a vertex. – Manuel Lafond Jun 23 '14 at 17:10
  • Isn't that the same as starting from a lower-vertex graph and using the inductive hypothesis to show that it's true for the next step? – linkhyrule5 Jun 23 '14 at 20:45
  • Anyway, the statement is something like "Given that each node of the graph is an independent random variable, show that a one-to-one correspondence exists between the product space of the variables and the space of permutations of values in the graph." That is, given a DAG of random variables, you can "pull out" the randomness and point to "the version of this graph where the random variables take on these values." – linkhyrule5 Jun 23 '14 at 21:49
  • I'm interested in this myself. I'm not sure, but if the DAG is finite, it should correspond to well-founded partial order (I think?), in which case it should be possible: (1) https://math.blogoverflow.com/2015/03/10/when-can-we-do-induction/ (2) https://math.stackexchange.com/questions/3155166/can-we-expand-induction-principle-to-a-partial-order-x-leq Of course I would love to be able to see an example of this to make sure I'm not misunderstanding or misusring this. But anyway hopefully this could be useful to you or anyone else who sees this question after a google search in the future... – hasManyStupidQuestions Nov 10 '21 at 00:26

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