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I'm studying cs in germany, I have been going through my script starting with propositional logic. I don't really understand the difference between those. I watched some YT-Videos on the topic but i didn't really helped me differentiate between those. They just sometimes use one or the other.

"Definition" in my own words (my understanding):

  • The script says something like "$A ⇔ B$" gets when "$A ↔ B$" is true. So only if both statements A, B are true or both are false "$A ⇔ B$" would get used.
  • The Term "$A ≡ B$" gets used if two logical Terms (consisting of Variables, Constants, ...) result in a equal logical value.

Example:
$A ↔ B ⇔ (A → B) ∧ (B → A)$
$A ∧ (B ∧ C) ≡ (A ∧ B) ∧ C$.

Problem:When given to "equal" statements they will be right and wrong the same time therefore i can use $⇔$. But they also will have the same logical value so i can use $≡$. I can't really differentiate between them.
Almost a different question. Why do the de morgan laws use ↔, why don't the use one of $≡, ⇔$? $¬(A ∧ B) ↔ (¬A ∨ ¬B)$. English is not my first language and i translated my question mostly by myself. Also its my first time asking a question here.

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    $\begingroup$ I've always used them as basically different symbols for the same concept. There might be some reasons to distinguish them, but I have never needed them to be distinguished. I guess you could define $a\iff b$ as $(a\to b)\land (b\to a),$ while we could define $a\equiv b$ as $(a\land b)\lor(\lnot a\land\lnot b).$ Maybe it requires excluded middle to prove these are equivalent? $\endgroup$ Commented Jun 13 at 20:48
  • $\begingroup$ Observe that your bullet point summary of ⇔ ↔ ≡ actually says that they are thoroughly interchangeable. They do have different meanings, but even when this is respected, frequently the choice between them is a matter of emphasis anyway. See whether the links in this answer's addendum help make things a bit clearer. $\endgroup$
    – ryang
    Commented Jun 14 at 5:23

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It will be question of convention on notation, if you want to distinguish two operators, one can set the following.

Indeed here we can note two things :

Preliminary before math

Take two sentences :

  • $X=$ "The first day of the year is 1st January"
  • $Y=$ "The Earth is a planet"

Both $X$ and $Y$ are sentences. They have both a meaning different and independant of each other. And they are both true.

You see here, that we cannot have $X \implies Y$ or $Y \implies X$, it doesn't make any sense. It has no meaning. Recall that we use the notation $X ⇔ Y$ to say ($X \implies Y$ and $Y \implies X$).

So because there are not logical implication (very concrete in very real life), between $X$ and $Y$. So they can't be equivalent (so "$X ⇔ Y$" is not true). Do you agree with that concrete example ?

Yet both $X$ and $Y$ are both true,the first day of the year is 1st January and the Earth is a planet :

So $$ X \equiv Y$$

Meaning is always about a system, a language, which has a real, a concret meaning. Sometimes this language is English, French, German, sometimes its mathematic. Yet, in every language something can be true or can be false (or neither true nor false).

First

$$ ⇔ \text{and} \leftrightarrow $$

are the same object just noted differently.

When you note $A ⇔ B$ it means they have not only the same logical value but also the same meaning.

For example : $$x\in[a,b] ⇔ a \leq x \leq b$$

Secondly

The sign $\equiv$ is just here to say that thing have the same logical value (in binary logic, True or False).

For example :

$$ (\forall \in \mathbb{N}^*, n>1, n-1>0) \equiv (\pi=\pi) $$

Both are true, but have of course not the same meaning.

Moreover

You see that :

$$A= \ "[(\forall \in \mathbb{N}^*, n>1, n-1>0) ⇔ (\pi=\pi)]" $$

is false.

Yet :

$$ B= \ "[(\forall \in \mathbb{N}^*, n>1, n-1>0) \equiv (\pi=\pi)]" $$

is true.

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  • $\begingroup$ Can you find a case where the distinction is a real difference? $\endgroup$ Commented Jun 13 at 20:50
  • $\begingroup$ Between what and what ? $\endgroup$
    – EDX
    Commented Jun 13 at 20:50
  • $\begingroup$ A case where only one of $P\iff Q$ and $P\equiv Q$ is true? $\endgroup$ Commented Jun 13 at 20:52
  • $\begingroup$ Ok get your question, yes. $\endgroup$
    – EDX
    Commented Jun 13 at 20:53
  • $\begingroup$ you loose me at your second example. Whats the difference between meaning and logical value? Im not saying its wrong... $\endgroup$ Commented Jun 13 at 21:03

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