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In the comments Daniela mentions that the goal is to show that indeed the successor function is injective on $\omega$. As Hagen shows, if we take an arbitrary inductive set then the axiom of regularity is needed. But if we reduce to linearly ordered inductive sets, i.e. $X$ is inductive and for every $x,y\in X$ either $x=y$ or $x\in y$ or $y\in x$, then we ...

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If we abandon regularity, we may indeed face some weird sets $a\ne b$ such that $a=\{b\}$ and $b=\{a\}$. Then if $a,b\in X$ (e.g. $X=\omega\cup\{a,b\}$), injectivity fails indeed. But if you want to work only with the case $X=\omega$ (the smallest infinite ordinal), all is fine: Ordinals are regular automatically.

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Here's what Euclid had to say on the matter. Ostensibly, an equivalent sort of argument can be made as follows: We note that for any two circles, an inscribed square has perimeter proportional to the diameter of the circle. Similarly, we can show (inductively?) that an inscribed regular $2^{n}$-gon has perimeter proportional to the diameter of the circle. ...

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It is somewhat of a delicate matter. Firstly, the independence of the circumference-diameter ratio from the radius of the circle is not true in all geometries. It is in fact a characteristic of 'flat' geometries, or, to use the standard term, of Euclidean geometries. It is quite easy to see for instance that this ratio is not a constant for circles on a ...

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It can be proved that the ratio $\frac{\text{circumference}}{\text{diameter}}$ is the same for all circles, no matter how large or small; in other words, this ratio is a constant number. The symbol $\pi$ is used to denote this constant number. We need a special symbol because there's no other good way to write the number, really, since it is transcendental. ...

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One possible system that you might use for explaining a system of axioms without having to spend a huge amount of time might be Douglas Hofstadter's "MU" system. The ideas are explained here on Wikipedia, where it is described as a puzzle, but it could be presented as a system of axioms. You are initially given a string "MI" and a number of rules (axioms) ...

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Regarding the second question, "Is there any first order theory which is consistent by finitary proofs but inconsistent with infinitary proofs?" The answer is yes, when particular additional infinitary inference rules are used. Lord_Farin's answer shows that infinitary inference rules are necessary to make this happen. It is well known that the set $X$ of ...

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Yes, of course. We might need to "go one level higher" so to speak, and base our discussion in some given model of set theory (instead of on the metalanguage, where the behaviour of infinite objects is more shaky), but there are no real problems. Barring the introduction of new, infinitary proof rules, the answer is no. For, we have: Let $\Sigma\vdash ... 1 From my personal experience, an historical approach can be useful. Why not try with Morris Kline, Mathematics: The Loss of Certainty, Oxford University Press, 1980 . It explains the road to modern math, including the big isue regarding Foundations of Math (and the birth of Mathematical Logic). For an understanding of Math Log and Set Theory, I would ... 2 You seem to have misunderstood Godel's theorem. By Godel's theorems we know that$\operatorname{Th}(\mathbb{N},+,.,0,S)$is not recursively axiomatizable. But this does not at all imply that it is inconsistent. In fact it is consistent, since the theory has a model, namely$(\mathbb{N},+,.,0,S)$. 2 You forgot a hypothesis: a theory extending FOA can be both complete and consistent, if it is not axiomatizable by a recursively enumerable set of axioms. (roughly speaking, this includes fintie sets of axioms, as well as infinite sets of axioms so long as there is a "computer" program that can tell whether any given statement is one of the axioms or not) ... 1 (If I understood the question correctly:) Essentially, yes. You must make some unverifiable assumptions - i.e. assume some 'basic truths'. This is a big topic in the philosophy of mathematics - the most relevant reading on the subject would be on Goedel's Incompleteness Theorems. You may also find the idea of the Muenchhausen Trilemma relevant, which deals ... 0 Cut-the-Knot's SSS proof page has a number of solutions, including Euclid's. As the author indicates, however, only Hadamard's proof "goes through without a hitch", with the important aside: "assuming of course that isosceles triangles have been fairly treated previously". I'll give a full development of Hadamard's argument, including the necessary bits ... 1 As a direct answer, calculus has no axioms inherent to itself. Theorems of calculus derive from the axioms of the real, rational, integer, and natural number systems, as well as set theory. Most disciplines of modern mathematics exhibit this sort of behavior, in which the discipline has no axioms inherent to itself. Modern disciplines of mathematics ... 1 Assume the triangles are$ABC$and$A'B'C'$with sides$a,b,c$, and$a',b',c'$. First move vertex$A$to vertex$A'$(always possible). Then rotate to make coincide the sides$b$and$b'$(possible by assumption). The sides$a$and$a'$now part from the same point and are equal. So, they are radii of a circle with center at$C=C'$. A similar thing happens ... 3 How about a sequent calculus, with standard rules: $$\frac{\Gamma, P, P\vdash Q}{\Gamma, P\vdash Q}CL \qquad \frac{}{\Gamma,P \vdash P}Ax \qquad \frac{}{\Gamma, \bot \vdash P}{\bot}L$$$$\frac{\Gamma\vdash Q\quad \Gamma,P\vdash R}{\Gamma,P\leftrightarrow Q\vdash R}{\leftrightarrow}L_1 \quad \frac{\Gamma\vdash P\quad \Gamma,Q\vdash ... 4 As far as I know, this is an open problem. It might be the case that even$ZFC\$ axioms are not sufficient to settle all such questions, or that the problem is even undecidable (although it seems unlikely). The main issue that we know very little about behavior of repeated exponentiation. For example, it is unknown if the following number is an integer: ...

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I will answer negatively for consistent theories. My reasoning for this is based on an answer I gave to myself for a question with which I received no assistance. As part of that answer, I had to address the nature of defined terms with respect to consistent theories in first-order logic with identity. There are two issues here. One is understanding why ...

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