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Consider the statement,

$1.$ "If it is Tuesday, then it is raining".

In propositional logic, 1 would read as, "$p \implies q$." Now, in accordance with the rules and definitions prescribed in logic, we have a plethora of logical equivalences. We can rewrite 1 as $ \neg p \vee q$, and in English,

$2.$ "It is not Tuesday or it is raining."

By setting up a truth table, we can prove that these statements are equivalent, hence, 1 and 2 have the same meaning. However, if I were read the two statements in English, I wouldn't suspect that they have the same meaning in everyday language. My question is: does the fact that the two statements in logic have the same meaning necessarily imply that the have the same meaning in everyday language? Because I honestly don't see how they convey the same meaning.

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    $\begingroup$ Something in logic implies something in everyday language? Not in English, nor most other languages. Maybe in Esperanto or something... $\endgroup$ – GEdgar Jan 26 '13 at 21:13
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Technically speaking no they do not have to be equivalent. In your particular case it depends on the interpretation given to the "or" conjunctive in the English Language. Many people (and I am not saying it is wrong or right) interpret the statement "$A$ or $B$" as "$A$ is true, or $B$ is true, but not both at the same time". Using this interpretation, both statements are not equal. Nevertheless if we all agreed that "$A$ or $B$" means "$A$ is true, or $B$ is true, and it is possible that both are true" then the statements would be equal since the interpretation of $\vee$ and "or" would then be the same.

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While it’s perhaps not immediately evident, the statement It is not Tuesday or it is raining does at least imply the statement If it is Tuesday, then it is raining even in everyday language: if the first statement is true, and if today really is Tuesday, then it must in fact be raining. The second probably does not imply the first in most people’s everyday language, because in everyday usage if ... then is normally taken to imply some connection $-$ perhaps not a truly causal connection, but something along those lines. Material implication ($\to$) is definitely not the same as everyday if ... then, so a formal equivalence between statements involving material implication may not translate into an equivalence between the apparent everyday counterparts.

The same problem arises to some degree with disjunction ($\lor$). In everyday usage the word or is often closer to exclusive or ($\veebar$) than to ordinary inclusive $\lor$, especially when preceded by either: It was either John or Charles; Apologize to your sister, or leave the table!

More generally, there’s a problem translating between formal and everyday language. Consider the statement Touch me, and you’ll lose some teeth! On the face of it that is $p\land q$, where $p$ is you touch me, and $q$ is you will lose some teeth, but the purely formal version closest to the real sense is $p\to q$.

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  • $\begingroup$ The one directional implication from $\lnot p\lor q$ to $p\to q$ is unusually similar to the same relation in intuitionistic logic. I'm not certain whether english would consider the law of excluded middle to be the same as tautology, though. (The only circumstance I can see it arising is in an informal logical proof in conversation, if one is trying to reason about some event.) $\endgroup$ – Mario Carneiro Mar 6 '16 at 21:44
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It is strictly by definition, $p\Rightarrow q$ is true if and only if the consequence $q$ is true, or the antecedent $p$ is false. You can see it in the truth table that defines the implication. That is, $p\Rightarrow q$ is true if and only if either $\neg p$ is true, or $q$ is true; i.e., if and only if $\neg p\lor q$ is true..

Or you can simply look at the truth tables. The truth table of $\neg p\lor q$ is the same as the truth table of $p\Rightarrow q$: true if $p$ and $q$ are false; true if $p$ is false and $q$ is true; false if $p$ is true and $q$ is false; true if $p$ and $q$ are both true: $$\begin{array}{c|c||c} p & q & p\Rightarrow q\\ \hline 0 & 0 & 1\\ 0 & 1 & 1\\ 1 & 0 & 0\\ 1 & 1 & 1 \end{array}\qquad\qquad \begin{array}{c|c|c|c} p & q & \neg p & \neg p\lor q\\ \hline 0 & 0 & 1 & 1\\ 0 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 1 \end{array}.$$ The final columns are identical, so the two formulas take the same truth values given the same truth inputs: that is, they are propositionally equivalent.


This may seem counter-intuitive, particularly with respect what is sometimes meant by "or" in standard usage: sometimes we speak of "this or that" when we really mean to say, "this or that, and not both". Logically, though, "or" is taken to mean "this or that or possibly both."

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