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Consider the following image: The axioms are, as Stephen Douglas Allen said: The first in a line of logic, just look at the picture. It means that they are assumptions made beforehand, and theorems are consequences of these assumptions (See Thm, 1.1) and consequences of other theorems (See Thm, 3.2) or consequences of theorems and axioms (See Thm, ...


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I quote myself from here: In my opinion, we should try to refrain from speaking of "axioms", since the term is basically meaningless. Admittedly, I tend to violate this recommendation all the time. Anyway, the way I see it, there are sentences. A collection of sentences can entail another sentence. If we have a collection $S$ of sentences, we can ...


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As axiom is an assumption we make that we consider to be true. That is, we decide that it is true. Because of this, an axiom is unprovable. It is true because we say it is true. All other laws, theorems, etc must be proven from the base set of axioms. An axiom can simply be a definition or it can be a theorem. A definition only identifies something, gives ...


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The set $A=\{A, \emptyset\}$ itself does not violate Regularity, since indeed there exists an element inside $A$ whose intersection with $A$ is empty. $$\emptyset\cap A = \emptyset$$ (Incidentally this property holds true for any set $A$.) However the set $\{A\}$ does violate Regularity: every element (there is only one) inside $\{A\}$ intersects $\{A\}$ in ...


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The other answers and the various comments clarified this important point. There are two senses of the expression (a) 'A violates (R)'. First, (a) may mean: if you let x=A in (R), then you get a false sentence. This is not true: A does not violate (R) in this sense. Second, (a) may mean: if you assume that A exists (plus the axioms of ZFC minus (R)), ...


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Let's suppose that there is a set $A$ such that $$A=\{A,\emptyset\}.$$ Now, consider the set $$B:=\{A,A\}.$$ This set exists by Pairing (if $A$ exists), and we can show by Extensionality that $$B=\{A\}.$$ Now we apply Regularity to get our contradiction. You have correctly observed that such a set $A$ does not violate Regularity. However, Regularity is not ...


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Suppose there were a set $A$ with the property $A=\{A,\emptyset\}$. Clearly $A \not = \emptyset$ since $A$ would have at least one element and possibly two. Now let $x=\{A\}$. Clearly $x \not = \emptyset$ since $x$ would have exactly one element. So the axiom of regularity would imply $\exists y\,(y\in x\land x\cap y=\emptyset)$. Since $x=\{A\}$ would ...


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We prove the following version of the Claim: Theorem: For any set-like well-ordered class $\langle A,\prec\rangle$, there is a (unique) $F$ such that $F$ is an order isomorphism from $\langle\alpha,\in\rangle$ to $\langle B,\prec\rangle$ for some (class) ordinal $\alpha$ and some initial segment $B$ of $A$, and one of these three holds: $B=A$ ...


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There is a counterexample to Goldbach somewhere between $10^{99999}$ and $10^{999999}$. There is an odd number of twin primes in the same interval. To prove or disprove just check all cases.


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The main focus on one or the other correspond to two different historical approaches of logic: In Frege's Begriffsschrift (1879), for example, logic is thought as a science, that is, a body of laws governing the notion of the Truth. Therefore, what is important for him is to hightlight the fundamental principles of thought and put them together, as the ...


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A logical axiom can be be considered a rule of inference that happens to have with no antecedents. Any interesting proof system must have at least one axiom (otherwise there is no way to begin a proof in the empty theory), but it must also have at least one rule of inference with premises (otherwise all you could prove would be the axioms themselves). ...


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In a logical approach: Assume we have a theory $T$ and its set of axioms $\Gamma$, such that $T=\{\varphi|\Gamma \vdash \varphi\}$ (a theory is closed under logical consequence). Now, for some reason, we want to postulate a new axiom to $T$, call it $\psi$. Clearly, we have a new axiom set $\Gamma'=\Gamma \cup \{\psi\}$ and so a new theory ...


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First question The point of induction is to convert the fact that you are able to prove $P(n)$ for any natural number $n$ into a proof that $P(n)$ for any natural number $n$. The two are 'very' different. The first is: For any natural number ( There is a proof that ( ... ) ). The second is: There is a proof that ( For any natural number ( ... ) ...


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Let's take an example: the axioms of a group. Now we wish to declare a new axiom: all elements commute. How do we establish that this new axiom is not a consequence of the old ones? By constructing a noncommutative group. How do we establish that it is not in contradiction to the original ones? By constructing a commutative group. Here are a couple of ...


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Pairing does not follow from (class) comprehension: the axiom of class comprehension allows us to do the dangerous sorts of operations which initiated the study of axiomatic set theory. For example, it allows us to create the class consisting of all sets. In order to avoid contradictions, classes are not allowed to be members of things. Your derivation uses ...


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NBG doesn't usually (at least as developed in e.g. Mendelson, Introduction to Mathematical Logic) contain one axiom scheme of class comprehension, but a finite set of axioms that together combine to a class existence metatheorem. However, those class existence axioms depend on the fact that we already know that ordered pairs exist for every pair of sets, ...



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