21
$\begingroup$

If $P$ and $Q$ are statements,

$P \iff Q$

and

The following are equivalent:

$(\text{i}) \ P$

$(\text{ii}) \ Q$

Is there a difference between the two? I ask because formulations of certain theorems (such as Heine-Borel) use the latter, while others use the former. Is it simply out of convention or "etiquette" that one formulation is preferred? Or is there something deeper? Thanks!

$\endgroup$
1
  • 51
    $\begingroup$ TFAE: (1) A iff B; (2) TFAE: (i) A; (ii) B. $\endgroup$ Jul 10, 2013 at 6:21

3 Answers 3

40
$\begingroup$

As Brian M. Scott explains, they are logically equivalent.

However, in principle, the expression $$(*) \qquad A \Leftrightarrow B \Leftrightarrow C$$ is ambiguous. It could mean either of the following.

  1. $(A \Leftrightarrow B) \wedge (B \Leftrightarrow C)$

  2. $(A \Leftrightarrow B) \Leftrightarrow C$

These are not equivalent; in particular, (1) means that each of $A,B$ and $C$ have the same truthvalue, whereas (2) means that either precisely $1$ of them is true, or else all $3$ of them are true. Also, you can check for yourself that, perhaps surprisingly, the $\Leftrightarrow$ operation actually associative! That is, the following are equivalent:

  • $(A \Leftrightarrow B) \Leftrightarrow C$
  • $A \Leftrightarrow (B \Leftrightarrow C)$.

In practice, however, (1) is almost always the intended meaning.

$\endgroup$
13
  • 2
    $\begingroup$ I would never write the second one without parentheses. Also because there's also a third possible interpretation: $A\iff(B\iff C)$. As a general rule, for a nested binary operator $@$, parentheses should only be omitted iff $(A @ B) @ C$ and $A @ (B @ C)$ are equivalent. $\endgroup$
    – celtschk
    Jul 10, 2013 at 7:04
  • 4
    $\begingroup$ @celtschk, biconditional is associative - see the last sentence of my answer. $\endgroup$ Jul 10, 2013 at 7:09
  • 3
    $\begingroup$ Ah, I missed that. That's indeed surprising. Although on second thought, it's perhaps not that surprising; after all, it makes sense that equivalence is an equivalence relation :-) $\endgroup$
    – celtschk
    Jul 10, 2013 at 7:15
  • 1
    $\begingroup$ Well, just ignore that remark; I was temporarily confusing associativity and transitivity. I shouldn't write comments before the second coffee, I guess ;-) $\endgroup$
    – celtschk
    Jul 10, 2013 at 9:15
  • 1
    $\begingroup$ Actually I now noticed that in three-valued Łukasiewicz logic they are not equivalent: $((F\iff U)\iff U) = (U\iff U) = T$, but $(F\iff(U\iff U))=(F\iff T)=F$ $\endgroup$
    – celtschk
    Jul 11, 2013 at 9:14
21
$\begingroup$

They are exactly equivalent. There may be a pragmatic difference in their use: when $P$ and $Q$ are relatively long or complex statements, the second formulation is probably easier to read.

$\endgroup$
1
  • 18
    $\begingroup$ Also, TFAE is very nice for when there is more than one claim (especially when there is no obviously preferable way of ordering them in an IFF chain). $\endgroup$
    – anon
    Jul 10, 2013 at 6:05
0
$\begingroup$

"TFAE" is appropriate when one is listing optional replacements for some theory. For example, you could list dozen replacements for the statements, such as replacements for the fifth postulate in euclidean geometry.

"IFF" is one of the implications of "TFAE", although it as $P \rightarrow Q \rightarrow R \rightarrow P $, which equates to an iff relation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .