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A definition of an object \(x\) is called impredicative if the expression used to explain the value of \(x\) contains mention of \(x\) itself. The example that I want to ask about is perhaps illustrative enough: In Group Theory, \(e\) is a neutral element for \(G\) if \(ex = xe = x\) holds for every element \(x\) of \(G\). Since \(e\) is itself an element of \(G\), this is a typical example of an impredicative definition. This statement is the axiom that a group has a neutral element. Therefore I began wondering if perhaps Group Theory is an impredicative theory. My original motivation comes from a nice booklet written by Edward Nelson titled Predicative Arithmetic: web.math.princeton.edu/~nelson/books/pa.pdf The title of Chapter 1, which has a length of about one printed page, is "The impredicativity of induction". Superficially a single axiom of the axiom schema of induction appears to have a similar structure to the example from Group Theory. If \(P\) is a predicate for which both \(P(0)\) and \(P(n) \rightarrow P(n+1)\) hold, then \(P(n)\) is true for all \(n.\) This axiom does again contain a universal quantifier. But if you look for an element which is defined and which is also similarly quantified by the quantifier, you are harder pressed than in the group example above, at least in my limited understanding. Nelson must have meant something by this. Unfortunately he passed on some years ago. Maybe someone can provide an insight? I honestly believe that my understanding is not up to par with this. The point of it all is that a theory that contains impredicative definitions is somehow reasonably deemed deficient. A little similar to having circular definitions. I did ask around for an explanation for why Group Theory is not equally considered deficient. Now I am not mentioning names, but I got the explanation from a very knowledgeable mathematician that the point is that you may substitute the above definition of the neutral element \(e\) by the equivalent definition "there exists a unique element \(e\) in \(G\) with the property \(ee = e.\)" This is indeed the case. The problem now is that to say the \(e\) is `unique' means to say \(\forall x : xx = x \rightarrow x = e,\) and now you have a universal quantifier which has \(e\) in its scope, and it makes the definition impredicative. If you do not include the `unique' part, I leave it as an exercise to show that \((\{ 0,1 \},\cdot)\) becomes a group, where \(\cdot\) is usual multiplication. My question here is whether there is some clear argument to say that the definition of a group is not necessarily impredicative?

Lol considering I am only 5 foot seven and 155 lbs my legs can press 250 kg for ten reps on a Universal machine. 60 kg is nothing.