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Are all definitions or axioms in logic biconditional (iff) statements? It would make sense to me that they would be. A lot of times I will read a definition though and it won't be written as an iff. Just curious as to why they are or aren't.

W. G.
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  • I should say particularly in propositional calculus the $\iff$ – W. G. Aug 26 '17 at 20:14
  • Axioms need not be. Definitions are always of the form: When we use this word we mean precisely these conditions. If the conditions, then we use the word. If we use the word, then those conditions. Hence, the word iff the conditions. So definitions are if and only if statements. However in my opinion it seems very artificial and heavy handed to present them as such unless one is using a purely logic framework. Otherwise assignment and identity seem more fundamental to me then logical conditionals. – fleablood Aug 26 '17 at 20:15
  • Why wouldn't axioms be then? I mean I read a post from awhile back that they are treated which seemed to be identically from a logic stand point. – W. G. Aug 26 '17 at 20:21
  • Why would they be? An axiom is any statement taken to be without proof. The statement could be ... anything. So if my axiom is "all dogs are fish" that's not an if and only if statement. – fleablood Aug 26 '17 at 20:26
  • Axioms are list of things we want our objects to satisfy. There is absolutely no reason to require that they are biconditional. When somebody asks me what kind of coffee I want, I usually say: 1) I want black coffee. 2) I want no sugar. These are the two axioms of "coffee I like". Notice that they are not biconditional, yet perfectly describe what I want. Could you give example of axioms that are in form of biconditional and some axiom that is not, but you feel should be? – Ennar Aug 26 '17 at 20:32
  • For example, couldn't you rewrite the axiom of exstensionality as: For sets x and y we say $x=y$ iff $\forall x \forall y ( \forall z (z\in x ...$. I guess I was just curious why I don't see iff statements after all my definitions. – W. G. Aug 26 '17 at 20:32
  • Axiom $\neq$ definition. – Ennar Aug 26 '17 at 20:36
  • Axiom of exstensionality is an axiom. I would not call it a definition though, because that is what it is known as. What is the difference between an axiom and a definition then? I remember reading another post asked awhile ago about this and it said nothing from a logic standpoint. – W. G. Aug 26 '17 at 20:41
  • An arbitrary axiom $\phi$ is equivalent to the biconditional $\phi \Leftrightarrow (x = x)$. (However, I think the questioner is really asking the question that has been nicely answered by Will Hunting.) – Rob Arthan Aug 26 '17 at 20:53
  • Axioms are formulas that some objects satisfy and some don't. For example, commutativity is an axiom that some groups satisfy and others don't. Definitions are statements that give name to objects satisfying a list of axioms. Formulas are not required to be biconditional as that would be too restrictive. Definitions need to be biconditional because you want to be able to freely exchange between terms and list of axioms they satisfy. – Ennar Aug 26 '17 at 20:56
  • @Enner: any first-order formula is equivalent to a biconditional, see my comment. – Rob Arthan Aug 26 '17 at 22:56
  • Consider the statement:

    $$\forall x,y \in N: x+y\in N \land \forall x\in N:x+0=x \land \forall x, y\in N: x+S(y)=S(x+y)$$

    No biconditionals. Does this not define $+$ on the set of natural numbers $N$?

     – Dan Christensen Aug 27 '17 at 04:51

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Definitions in mathematics always go in both directions. For example, when we say we define $X$ to be a Riemann surface if it is a one-dimensional complex manifold, we really mean iff here. However, it is traditional in English to write definitions using if instead of iff, and most books still use if, though there are some which use iff. The same goes for axioms if we are defining something using axioms. For example, when we say we define $G$ to be a group if it satisfies the group axioms, we really also mean iff here.

  • However, individual axioms my indeed be conditionals, rather than biconditionals. – Jim H Aug 26 '17 at 22:32
  • Nice answer but did you mean two-dimensional instead of one-dimensional? – Giorgio Mossa Aug 27 '17 at 08:08
  • @WillHunting Ahhhhh I see, so you were talking of holomorphic manifolds. Thanks for clarifying. ;-) – Giorgio Mossa Aug 27 '17 at 12:34
  • Of course, definitions are not "really" biconditional in the sense that, apart from the definition itself, there is no way to prove (for example) that a Riemann surface is a one-dimensional complex manifold. Both "if" and "iff" in definitions are abbreviations for something other than a conditional or biconditional, although in English it can appear similar. – Carl Mummert Aug 28 '17 at 12:17
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Definitions, yes. Axioms not necessarily.

Note that when we are looking for a definition of $X$, we ask: "What are the necessary and sufficient conditions for something to be an $X$?"

If $P$ is a sufficient condition for $Q$, we write $P \rightarrow Q$

If $P$ is a necessary condition for $Q$, we write $Q \rightarrow P$

So, if $P$ is a necessary and sufficient condition for $Q$, we get $(P \rightarrow Q) \land (Q \rightarrow P)$, which is of course just $P \leftrightarrow Q$

So that's where the biconditional comes from in case of definitions.

Axioms can be used to capture definitions as well, in which case we call the definitional axioms. But not all axioms express definitions.

Bram28
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