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I am confused about the difference between ↔ (biconditional iff) and ≡ (logical equivalence). For instance, p→q can be rewritten as ∼p∨q. Would it be correct to say p→q↔∼p∨q or p→q≡∼p∨q?

Secondly, is ⇔ another symbol for ≡?

Finally, what's the difference between → and ⇒?

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    $\begingroup$ You should look in that book to see what the symbols mean. For example, it could be that $\rightarrow$ combines two wffs into a new wff, while $\Rightarrow$ is a relation between wffs. $\endgroup$
    – GEdgar
    Sep 17, 2017 at 0:31
  • $\begingroup$ The issue is not with the symbols but with the concepts. We have a connective: the biconditional that can produce a "complex" sentence (or formula) from simpler ones: $p \leftrightarrow q$. $\endgroup$ Sep 18, 2017 at 6:54
  • $\begingroup$ And we have the relation of logical (or semantical) equivalence between formulas. Logical equivalence is different from the biconditional, although the two concepts are closely related; in a nutshell: $p \leftrightarrow q$ is a tautology iff $p$ is logically equivalent to $q$. $\endgroup$ Sep 18, 2017 at 6:57
  • $\begingroup$ See also this answer on SE Philosophy: philosophy.stackexchange.com/questions/65024/… $\endgroup$ Aug 30, 2019 at 18:38
  • $\begingroup$ en.wikipedia.org/wiki/Logical_connective#History_of_notations seems to suggest they could mean the same across time $\endgroup$
    – Make42
    May 16, 2022 at 14:48

3 Answers 3

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In short, $P \leftrightarrow Q$ is statement that could be either true or false. $P \equiv Q$ means that $P \leftrightarrow Q$ is always a true biconditional (so, $P$ and $Q$ have the same truth value no matter what).

So, one could say that $\neg (P \vee Q) \equiv \neg P \wedge \neg Q$ (DeMorgan's) but you typically wouldn't write $\neg (P \vee Q) \leftrightarrow (\neg P \wedge \neg Q)$.

The arrow $\Rightarrow$ usually is slang for "implies" but different people use it differently. The arrow $\Leftrightarrow$ is usually treated the same way as $\leftrightarrow$.

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  • $\begingroup$ I found it's clearer to mention the components in logical equivalence: $\phi \equiv \psi$ means for that every truth assignment applied to the components of $\phi$ and $\psi$, the resulting truth values of $\phi$ and $\psi$ are identical. $\endgroup$
    – Tran Khanh
    Oct 24, 2023 at 4:15
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In case you can use a somewhat philosophical explanation: $\leftrightarrow$ is a logical operator within statements, while $\equiv$ serves to state an equivalence between statements and thus may be thought of as meta-logical.

As Randall explained, $P \leftrightarrow Q$ is a statement $-$ one statement and a logical statement. The $\leftrightarrow$ will cause it to be true under certain truth value distributions for $P$ and $Q$. The same applies to $\neg(P \vee Q) \leftrightarrow (\neg P \wedge \neg Q)$.

For $\neg(P \vee Q) \equiv (\neg P \wedge \neg Q)$ however, you compare two truth tables, the one of $\neg(P \vee Q)$ and $(\neg P \wedge \neg Q)$, two distinct statements. If and only if both are true exclusively for the same truth value distributions, the equivalence applies and so the meta-logical statement $\neg(P \vee Q) \equiv (\neg P \wedge \neg Q)$ is true.

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    $\begingroup$ This is what I've expected to hear actually. So thank you for the insight! $\endgroup$ Sep 20, 2019 at 12:56
  • $\begingroup$ A fancy way to put this is to say that $\leftrightarrow$ appears in object-language whereas $\equiv$ appears in meta-language $\endgroup$
    – KGhatak
    Jun 20, 2020 at 7:57
  • $\begingroup$ Do you mean that $\ldots≡\ldots$ is the same as $\forall \ldots ↔\ldots$? $\endgroup$
    – Make42
    May 16, 2022 at 14:45
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Consider this analogy. You wouldn't say the following:

Prove that 3 + 5.

You might instead say this:

Prove that 3 = 5.


Similarly, following doesn't make sense:

Prove that $p \leftrightarrow q$.

Instead one of these would be correct:

Prove that $p \leftrightarrow q$ is always true.
or
Prove that $p \equiv q$.

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