I just want to ask a very elementary question.  When we introduce a "definition" in a first order logical system. For example when we say   Define: $Empty(x) \iff \not \exists y (y \in x) $   Isn't that definition itself an "axiom", call it a definitional axiom.  I'm asking this because the one place predicate symbol Empty() is actually new, it is not among the listed primitives of say Zermelo, which has only identity and membership as primitive symbols.   So when we are stating definitions are we in effect stating axioms? but instead of being about characterizing a primitive, they are definitional axioms giving a complete reference to a specified set of symbols in the system.  Is that correct?  Now if that is the case, then why we don't call it axiom when we state it, I mean why we don't say for example:  Definitional axiom 1) $Empty(x) \iff \not \exists y (y \in x)$  Zuhair
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I am wondering what the difference between a definition and an axiom.  Isn't an axiom something what we define to be true?  For example, one of the axioms of Peano Arithmetic states that $\forall n:0\neq S(n)$, or in English, that zero isn't the successor of any natural number. Why can't we define 0 to be the natural number such that it isn't the successor of any natural number?  In general, why is an axiom not a definition?   Are there cases where we can formulate definitions as axioms or vice versa?
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"calculus" and "formal system"  From      a calculus is a formal system that consists of          a set of syntactic expressions (well-formed formulæ or wffs),    a distinguished subset of these expressions (axioms), plus    a set of formal rules that define a specific binary relation on the space of expressions.      From      Formal systems in mathematics consist of the following elements:         A finite set of symbols (i.e. the alphabet), that can be used for constructing formulas (i.e. finite strings of symbols).   A grammar, which tells how well-formed formulas (abbreviated wff) are constructed out of the symbols in the alphabet. It is   usually   required that there be a decision procedure for deciding whether a   formula is well formed or not.   A set of axioms or axiom schemata: each axiom must be a wff.   A set of inference rules.      what is the difference between a formal system, and a calculus, then? I think they are the same? "calculus" and "logic system"  "Calculus" appears in "propositional calculus" and "first-order predicate calculus", which are also called "propositional logic" and "first-order logic" respectively. So I thought "calculus" and "logic" mean the same, and "a logic" is, according to ,      A logical system or, for short, logic, is a formal system together with a form of semantics, usually in the form of model-theoretic   interpretation, which assigns truth values to sentences of the   formal language, that is, formulae that contain no free variables.    But then I saw "calculus" also appears in "lambda calculus", which is also a formal system. I think a lambda calculus isn't a logic system, right? What does "calculus" mean in "lambda calculus"? Furthermore  "calculus" can also mean computational real analysis for first-year college students.   Thanks.
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I have a doubt. I need help.  Can the basic axioms of equality be presented as "open axioms"?  I) (reflexivity) $\qquad x = x$              II)  (Substitutivity) $\qquad (x = y) \to \big(F (x, x) \to F (x, y)\big)$
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I am looking to study mathematical logic, however, I find that introductory books are very daunting, which kind of disheartens me. You see, slowly but surely, I started to realize that the maths which I have learned did not just pop out of thin air, but is a collection of systems, which must of been developed via some other system, i.e, maths did not develop itself.  So I began to look into the origins of mathematics, and read that it was developed via a type of logic, which exists sort of by 'default', via a set of axioms, and then of course I looked up the definition of axioms.   So given that I'd be studying a type of logic whose origins are self evident axioms, naturally I believed there would be no prerequisites. However, in looking up mathematical logic, I have come across things such as Boolean algebra, sets, first order logic, some other type of logic, called 'traditional logic', as well as references to a sort of calculus, though not in a mathematical sense, I think.  So all in all, I am trying to develop a type of mental spider web, and I am trying to find out the strands which lye at the absolute bounds so that I may learn this mystique logic. Though I have no idea where to start.  Side note: This is the book I have started reading:   Credit goes to Wolfgang Rautenberg.
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I don't know if this is appropriate for math.stackexchange, or whether philosophy.stackexchange would have been a better bet, but I'll post it here because the content is somewhat technical.  In ZFC, we can prove that the (second-order) Peano axioms have a model, call it $\mathbb{N} = (N,0,S)$. Furthermore, $\mathbb{N}$ is unique up to isomorphism. Thus, it would seem that we've pinned down $\mathbb{N}$.  However, if ZFC is consistent, then it has some very peculiar models. In particular, it has a model $(V_0,\in_0)$ whose "native" $\mathbb{N}$, let call it $\mathbb{N}_0$, is a model for the sentence "ZFC is consistent," and another model $(V_1,\in_1)$ whose "native" $\mathbb{N}$, lets call it $\mathbb{N}_1$, is a model for the sentence "ZFC is inconsistent."  But in reality, in other words, for the "real" $\mathbb{N}$, only one of these can be sentences can be true. Furthermore, the objects $\mathbb{N}_0$ and $\mathbb{N}_1$ aren't isomorphic, since the set of sentences they satisfy are different.  So it seems to me that $\mathbb{N}$ cannot truly be pinned down. Furthermore, its categoricity is, in some sense, illusory. Is this correct, or am I missing something?
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If the empty set is a subset of every set, why it isn't written with the elements of a set? like so $\{1,2,3,\emptyset\}$   Or why isn't $\{\emptyset,\{a\}\}=\{\{a\}\}$? I know one has two elements and the other has 1 but since the empty set is a subset of both, then why it isn't being mentioned explicitly in the definition of the set?
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I'm working with the universal and the existential quantifiers and I was wondering is there an order of operations that applies to them? What is the difference between   $(\forall x)P(x)(\exists x)Q(x)$  And  $(\exists x)Q(x)(\forall x)P(x)$  Lets say we have this relationship  (a) $[(\forall x)P(x) \vee (\exists x)Q(x)] \implies (\forall x)[P(x) \vee Q(x) ]$
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While proving a universal quantification, say $\forall a\in D,~P(x)\rightarrow Q(x)$. I see (at least) two type of people, one begin the proof by  (1) Way 1     Let $a$ in $D$,   (or let $a$ be any element in $D$)      $\cdots\cdots$ (some arguments)      $\therefore P(a)\rightarrow Q(a)$      Since we made no special assumptions on $a$, any $a$ in $D$ also deduce   $P(a)\rightarrow Q(a)$. Thus we have shown that $\forall a\in D,~P(x)\rightarrow Q(x)$. (This is exactly what the U.G. rule says.)   one by: (2) Way 2     Arbitrary choose $a$ in $D$,       $\cdots\cdots$ (some arguments)      $\therefore P(a)\rightarrow Q(a)$      Since what we choose is arbitrary, so every element in $D$ also holds true. (some blur here.. Even I guess whether it is "exactly" a usage of U.G. rule)   I'm not ask for a elementary problem that how to prove a universal quantification. Instead, I'm curious which formulation of such a proof is more proper, logical and formal logical structure.  I think way 2 is somewhat not clear since yes, we had indeed arbitrary chosen a element $a$ in $D$, but someone may intensionally argue that "hey, it is because  you're lucky that you randomly chose the right $a$ at the very begin that makes the predicate true in the set! If you want to show that this is true for any element in the set, you should not have use the word choose in the proof's begin; or at least, you should say at end that since our intermediate steps do not use the special assumption of $a$, so it suits for all elements. But in this way, this becomes slightly complex than way 1, why not just use way 1?"   The reason that way II is valid is somewhat of a meta-level, not indicate by the language itself.
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I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).  Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.  Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.  Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".  EDIT : How is the term "wrong" defined in mathematics then ? You can say that proving fermat's last theorem using the number-theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms .  Is mathematics as rigorous and as thought-through as it is believed and expected to be ? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals .
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I was wondering why do some people use redundant axioms in definitions?  If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free.   Why do we then assume it?   Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R.
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We define the term "Set" as,     A set is a collection of objects.   And an "Empty set" as,     An empty set is a set which contains nothing.    First problem I encountered:  How the definition of "Empty set" is consistent with the definition of "sets" if "Empty set" contains nothing and a "set" is a collection of objects.  Further we discovered in set theory that every set has a subset that is the Null set.  Such as,  If $A=\emptyset$ and $B=\{1,2,3\}$ then,  $$A \subset B$$  Second problem I encountered:   How and why "No element" is referred and considered as an element as we do in case of null set that is, when we say that every set has a subset that is the Null set?  Third and Last one:  How can a set possess "some thing" and "nothing" simultaneously that is when we say that every set (containing objects) has a subset which is Null set (contains nothing)?
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Thus, axioms are propositions we assume to be true and we use them in an axiomatic theory as premises to infer conclusions, which are called "theorems" of this theory.  For example, we can use the  to prove theorems of arithmetic.  This is one meaning of the word "axiom". But I recognized that the word "axiom" is also used in quite different contexts.  For example, a group is defined to be an algebraic structure consisting of a set $G$, an operation $G\times G\to G: (a, b)\mapsto ab$, an element $1\in G$ and a mapping $G\to G: a\mapsto a^{-1}$ such that the following conditions, the so-called group axioms, are satisfied:   $\forall a, b, c\in G.\ (ab)c=a(bc)$, $\forall a\in G.\ 1a=a=a1$ and $\forall a\in G.\ aa^{-1}=1=a^{-1}a$.      Why are these conditions (that an algebraic structure has to satisfy to be called a group) called axioms? What have these conditions to do with the word "axiom" in the sense specified above? I am really asking about this modern use of the word "axiom" in mathematical jargon. It would be very interesting to see how the modern use of the word "axiom" historically developed from the original meaning.   Now, let me give more details why it appears to me that the word is being used in two different meanings:  As peter.petrov did, one can argue that group theory is about the conclusions one can draw from the group axioms just as arithmetic is about the conclusions one can draw from the Peano axioms. But in my opinion there is a big difference: while arithmetic is really about natural numbers, the successor operation, addition, multiplication and the "less than" relation, group theory is not just about group elements, the group operation, the identity element and the inverse function. Group theory is rather about models of the group axioms. Thus: The axioms of group theory are not the group axioms, the axioms of group theory are the axioms of set theory.  Theorems of arithmetic can be formalized as sentences over the signature (a. k. a. language) $\{0, s, +, \cdot\}$, while theorems of group theory cannot always be formalized as sentences over the signature $\{\cdot, 1, ^{-1}\}$. Let me give an example: A typical theorem of arithmetic is the case $n=4$ of Fermat's last theorem. It can be formalized as follows over the signature $\{0, s, +, \cdot\}$: $$\neg\exists x\exists y\exists z(x\not = 0\land y\not = 0\land z\not = 0\quad\land\quad x\cdot x\cdot x\cdot x + y\cdot y\cdot y\cdot y = z\cdot z\cdot z\cdot z).$$ A typical theorem of group theory is  which states that for any finite group G, the order of every subgroup H of G divides the order of G. I think that one cannot formalize this theorem as a sentence over the "group theoretic" signature $\{\cdot, 1, ^{-1}\}$; or can one?
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In ZFC there seem to be two ways to define a function, as an ordered triple of domain, codomain and the set of ordered pairs that make its graph, or a relation over the two sets.   Whichever we use doesn't matter, but this seems to present a certain problem. Both are defined over sets, but no-one seems to have any problem defining functions on the ordinals like Hartog's function giving the minimum ordinal that doesn't have an injective function on to the given ordinal. I know we want to be able to obviously, but how are we supposed to justify this, because the definitions of functions don't seem to include classes. Or is there a nicer definition I've missed?
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In ZFC, for example, there is no universal set, so what does it mean to write $\forall x (\cdots)$, i.e., for all sets something is true? Does it avoid the problem by quantifying over all elements but not allowing those elements to form a set?  A related question I have is about quantifying over infinite sets. If $X$ is a finite set with $n$ elements $x_1, x_2, x_3, \ldots, x_n$, Then it is reasonable to define $\forall x\in X :P(x)$ to be the statement $P(x_1)\wedge P(x_2)\wedge\cdots\wedge P(x_n)$ (which can be formalized using a recursive definition). But what about infinite sets? It seems that if $Y$ is a countably infinite set with elements $y_1,y_2,\ldots$, then $\forall y\in Y: P(y)$ would mean $P(y_1)\wedge P(y_2)\wedge\cdots$, which we couldn't finish stating. For uncountable sets, we would miss an element in between every time we count upwards.
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Motivation: If operation is change, and there is no change, then there is no operation. Then the meaning of  1+0 =1 is not in the equality itself but in the comparison with other equalities as  1+2=3. The following definition does not use the meaningless equality  1+0 =1, or generally a*e=a. I guess that trying to prove statements using the next definition, matters/problems will arise that do not arise with the standard definition. Let be the non empty set G = {a,b,c,d,.....z} with the operation *, satisfy the following 3 conditions:  a(bc) = (ab)c, for every a, b, c ∈ G (associative law). For every a,b there is c (unique) such that ac = b. (c  is the &quot;path&quot; from a to b). For every a,b there is d (unique) such that bd = a. (d is the &quot;path&quot; back from b to a).  P.S.: the meaning of 2) and 3) is the possibility to &quot;go&quot;/&quot;move&quot; from any member to any member, through the same path (if a-&gt;b-&gt;c then c-&gt;b-&gt;a as in the Klein group) or through other path (if a-&gt;b-&gt;c then c-&gt;X-&gt;a). Prove: G,*  is a group. Proof: Associativity: From 1) Identity element: Let's prove for every a there is e such that ae=ea=a. Proof: According to 2) For every a,a there is e such that ae = a. (3) According to 2) for every e,a there is x such that ex=a, that means ae =ex =a (4) Let's proof x=a. Multiply by a on the left, such that the result is some g ∈ G: ae = ex = a (4) &lt;---&gt; (a)ae = (a)ex = aa =g &lt;---&gt; aae = (ae)x = aa =g (associative) &lt;---&gt; &lt;---&gt; aae = ax = aa =g (3). (5) According to 2) for every a,g there is x (unique) such that aa=g and ax=g , then a and x are the same, and it's proved x=a. Then replacing in (5): ae = ea = a. Inverse element: Let's prove that for every a, there is b such that a ⋅ b = e and b ⋅ a = e According to 2) for every a,e there is b such that ab = e. (6) According to 2) for every b,e there is X such that bX=e. (7) Let's proof that X=a ab = bX = e (6),(7) Multiply be a on the right: aba =ea a(ba) = ea; Associativity ea = a; Identity element a(ba) = a; then ba = e bX=e according to (7) ba = bX = e; according to 2) for every b,e there is x unique such that bX = e then x=a and it's proved that for every a, there is b such that a ⋅ b = e and b ⋅ a = e.
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Why does $\bigcup \emptyset = \emptyset$ but $\bigcap \emptyset$ is not defined?    If I had to guess, I'd say it's also equal to $\emptyset$.
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In Peano arithmetic (first order), we first define natural numbers using a successor function and Peano axioms, then we define addition (and multiplication), and then, we define inequality, as:  $a\leq b\leftrightarrow\exists c\left(a+c=b\right)$   Is there any way to define inequality first, directly from the successor function and Peano axioms? (I mean, if I don't need addition for my purpose, why define it?).