22
$\begingroup$

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
  • 48
    $\begingroup$ TFAE: (1) A iff B; (2) TFAE: (i) A; (ii) B. $\endgroup$ – Amit Kumar Gupta Jul 10 '13 at 6:21
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 '13 at 7:04
  • 4
    $\begingroup$ @celtschk, biconditional is associative - see the last sentence of my answer. $\endgroup$ – goblin GONE Jul 10 '13 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 '13 at 7:15
  • $\begingroup$ @celtschk, okay I've just reorganized a little bit to make things clearer. $\endgroup$ – goblin GONE Jul 10 '13 at 7:17
  • $\begingroup$ Yes, with that rearrangement, the statement is no longer easy to miss. $\endgroup$ – celtschk Jul 10 '13 at 7:19
22
$\begingroup$

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 '13 at 6:05
1
$\begingroup$

"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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.