0
$\begingroup$

I am having a hard time understanding this section in Wikipedia's article on Logical biconditionals:

Colloquial usage

One unambiguous way of stating a biconditional in plain English is of the form "b if a and a if b". Another is "a if and only if b". Slightly more formally, one could say "b implies a and a implies b". The plain English "if'" may sometimes be used as a biconditional. One must weigh context heavily.

For example, "I'll buy you a new wallet if you need one" may be meant as a biconditional, since the speaker doesn't intend a valid outcome to be buying the wallet whether or not the wallet is needed (as in a conditional). However, "it is cloudy if it is raining" is not meant as a biconditional, since it can be cloudy while not raining.

Should the example read:

"I'll buy you a new wallet if you need one" may be meant as a biconditional, since the speaker does intend a valid outcome to be buying the wallet whether or not the wallet is needed (as in a biconditional).

My question is how can the plain English "if'" sometimes be used as a biconditional? I'm OK with the word "biconditional." I don't understand how the reader is to know the "speaker doesn't intend a valid outcome to be buying the wallet whether or not the wallet is needed (as in a conditional)" especially how this amounts to "(as in a conditional)".

  • 4
    I'm not sure there is a _mathematical_ question here. It may belong better on English.SE?2012-07-13
  • 1
    @HenningMakholm As the wikipedia article states: "One unambiguous way of stating a biconditional in plain English is .." Is not stating conditions unambiguously in plain English of interest to students of math?2012-07-13
  • 2
    The interest of mathematics students is to express *mathematical* statements unambiguously. Luckily, real world objects such as wallets, money, and the acts of buying are not mathematical statements.2012-07-13
  • 0
    English is not math, and the word 'if' in English should not be taken to be equivalent to mathematical implication. For example, there's the biscuit conditional: "There's a biscuit on the counter, _if_ you're hungry." There's no conditional implied: it's there whether or not you're actually hungry.2016-05-16

3 Answers 3

1

I think the biconditional means:

Buy wallet $\iff$ you need the wallet

So, by taking the contrapositive,

Don't buy wallet $\iff$ You don't need the wallet.

Hence, Wikipedia means that the speaker doesn't intend a valid outcome to be buying the wallet whether or not the wallet is needed , since if the wallet is not needed, the outcome is "Don't buy wallet".

If it were a conditional, i.e. "You need wallet $\Rightarrow$ Buy wallet", note that if "You need wallet" is false, "Buy wallet" could still be true and satisfy the truth table of the conditional.

Hope it helps.

  • 0
    Thank you for your answer. Should the parenthetical statement in the article be changed from "(as in a conditional)" to (as in a biconditional)?2012-07-13
  • 1
    The original "(as in a conditional)" seems to be correct.2012-07-15
  • 0
    "...buying the wallet whether or not the wallet is needed" cannot be the "exclusive-if," but then the wiki example has the parenthetical statement: "(as in a conditional)" this is where I get confused. What is the parenthetical "conditional" referring to?2012-07-19
  • 0
    @skullpatrol: The parenthetical (as in a conditional) refers to buying the wallet whether or not the wallet is needed .2012-07-22
2

It is lucky that expressing a biconditional in a mathematical statement is easy; there is an unambiguous mathematical meaning for "if and only if". Perhaps the philosophy.SE would be more inclined to give you a more in-depth answer. Implication is a subject of interest there since it isn't always merely the material conditional, $P\rightarrow Q$ meaning $Q\vee \neg P$, but can also have modal quantifiers. That would be something like "Necessarily, P implies Q" vs. "P implies (necessarily Q)".

1

No. Let $Q$ be the statement "I will buy you a wallet" and $P$ the statement "You need a wallet. "I'll buy you a wallet if you need one" could conceivably be translated as $P \Rightarrow Q.$ However, I hope you would agree that buying a wallet for someone who doesn't need one is quite silly - thus buying someone a wallet implies that they had needed one, ie) $Q \Rightarrow P.$ Hence $(Q\Rightarrow P ) \wedge (P \Rightarrow Q),$ or $P \iff Q$

  • 0
    Thank you for the answer. Are you saying that " buying a wallet for someone who doesn't need one is quite silly" is the same as "the speaker doesn't intend a valid outcome to be buying the wallet whether or not the wallet is needed"? If so, the parenthetical statement in the article should be changed from "(as in a conditional)" to (as in a **biconditional**).2012-07-13
  • 1
    @Former_Math_Addict: the cited phrase is not beautiful, but it means that the dismissed scenario (namely buying whether or not) would be valid if (only) the conditional is claimed, but is not valid if the biconditional is claimed. So the parenthetical statement says what it intends to.2012-07-13
  • 0
    @MarcvanLeeuwen Thank you for clearing that up.2012-07-13