Thomas Klimpel
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 9h comment Does a $\Pi_2^0$ sentence becomes equivalent to a $\Pi_1^0$ sentence after it has been proven? @DavidC.Ullrich If that were true, then the answer to my question should be a resounding YES! But the example of Goodstein's theorem given in the question should make in clear that a $\Pi_2^0$ sentence can be true, without being equivalent to every other true sentence. (I actually hoped that Goodstein's theorem would be equivalent to (or at least imply) the consistency of PA, but I only found such a statement for the related Paris-Harrington theorem.) The case of the model existence theorem is more problematic, because it isn't really a statement about natural numbers. 17h asked Does a $\Pi_2^0$ sentence becomes equivalent to a $\Pi_1^0$ sentence after it has been proven? Mar 29 comment Geometric intuition behind the Lie bracket of vector fields @sifsa See drive.google.com/folderview?id=0B6x6GQ82vuH_YnE2STNKZXU1cUk page 19 at the bottom. The proof exploits a definition of the Lie bracket by its effect on a test function $\varphi$ via $L_v(L_w \varphi) - L_w(L_v \varphi) = L_{[v,w]}\varphi$. Then it is sufficient to know that $L_v \varphi$ transforms canonically... Mar 23 comment Are there any objects which aren't sets? The point of sets in set theory is that we can quantify (impredicatively) over sets, because they are assumed to exist as a completed whole. Even so you can explicitly name some classes (even using predicative quantification over other classes), they don't exist as a completed whole. So you cannot meaningfully quantify (impredicatively) over them. I explained the way they fail to exist here, but it is probably hard to understand despite the simple examples. Feb 19 awarded Popular Question Feb 8 comment How Graph Isomorphism is used to determine Graph Automorphism? It is good that you care about graph automorphisms. Any permutation is a product of transpositions, but it can happen that you have two transpositions which are not automorphisms of the graph, but their product might still be an automorphism of the graph. But it is fine for me, if you say that the transposition was just an example, and that the idea also works for permutations which are not transpositions. On an unrelated note: Did you notice that the text said: "continue till an isomorphism is found"? So it only computes a single representative. Can you work out why a single rep. is enough? Feb 8 comment How Graph Isomorphism is used to determine Graph Automorphism? "Now consider a transposition $\pi$ that moves $(i+1)$ th vertex to $j'$ th position where \$i