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I was searching for a Latex symbol that indicates $A \Rightarrow B$ and $A \not\Leftarrow B$ ($B$ if not only if $A$, $B$ ifnf $A$). I thought of using $A \Leftrightarrow B$ with the left arrow tick < crossed out. Since I did not find such a symbol:

Is there a Latex symbol for this?

How common or understandable is this symbol?

If it isn't common: How easily is it confused with the symbol $\not\Leftrightarrow$?


Update:

I need it for a sequence $A$ ifnf $B$ ifnf $C$ ifnf $D$, which I find more understandable than $A \Leftarrow B \Leftarrow C \Leftarrow D$ and $A \not\Rightarrow B \not\Rightarrow C \not\Rightarrow D$.

Of course I will prove both directions.

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    $\begingroup$ perhaps this is a matter of taste, but i think this is the sort of thing better indicated using words $\endgroup$ – citedcorpse Jul 31 '13 at 17:37
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    $\begingroup$ This is probably not appropriate for this venue, but I'll give you my two cents' worth as a teacher and longtime LaTeX user: it's better to be clear and write the two separate statements as you did here, rather invent a symbol none of us has ever seen, let alone used. :) $\endgroup$ – Ted Shifrin Jul 31 '13 at 17:38
  • $\begingroup$ Perhaps you should ask this in tex.stackexchange $\endgroup$ – BlackAdder Jul 31 '13 at 17:40
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    $\begingroup$ If there is a real need to invent a new symbol, you can use \stackrel to stack a symbol on top another: $$A \stackrel{\Rightarrow}{\not\Leftarrow} B$$ $\endgroup$ – achille hui Jul 31 '13 at 17:41
  • $\begingroup$ I recommend you read the comments in this question, on why looking for such symbols may not be desirable. $\endgroup$ – Andrés E. Caicedo Jul 31 '13 at 17:47
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When people assert implications, they often implicitly involve universal quantification. For example, "if $n$ is a prime number greater than $2$, then $n$ is odd" really means "for all integers $n$, if $\dots$." When one denies an implication, one includes the universal quantifier in the denial, so it becomes an existential quantifier. For example, if someone says that "$n$ is odd" doesn't imply "$n$ is a prime number greater than $2$", he normally means to deny that "for all $n$, if $n$ is odd then $n$ is a prime number greater than $2$"; equivalently, he means to assert that "there exists an odd $n$ that is not a prime number greater than $2$." (Recall from propositional logic that the negation of $B\implies A$ is equivalent to $B\land\neg A$.) So your proposed combined connective, for implication in one direction and denial of implication in the other direction, will implicitly quantify the variables partly with universal quantifiers and partly with existential ones. This looks to me like a recipe for confusion and therefore well worth avoiding.

If, by good fortune, your statements $A$ and $B$ don't involve variables, so these quantification issues don't arise, then there is a fairly easy answer to your question. As I said above, the negation of $B\implies A$ is equivalent to $B\land\neg A$. Furthermore, this formula already implies that $A\implies B$, so $$ (A\implies B)\land\neg(B\implies A) $$ is equivalent to $B\land\neg A$. But remember, this use of propositional logic is legitimate only if your $A$ and $B$ don't involve any variables that are implicitly quantified in your implications.

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  • $\begingroup$ I worked out $(A\Rightarrow B)\land \neg(B\Rightarrow A)$ with the classical rules and was somewhat surpised for a moment when $A\Rightarrow B$ canceled out completely... $\endgroup$ – Nikolaj-K Jul 31 '13 at 18:50
  • $\begingroup$ Great answer, especially "recipe for confusion". $\endgroup$ – DaveBall aka user750378 Jul 31 '13 at 19:06
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It is doubtful that a symbol exists; I do not believe it is common usage.

Note that your situation is equivalent to "$A$ implies $B$, but $B$ does not imply $A$". There are many, many situations in mathematics when this is the case. For instance:

Independent random variables have zero correleation coefficient, but a zero correlation coefficient does not imply that the random variables are independent.

In fact, the distinction that $A \implies B$ does not imply that $B \implies A$ is so important that it is almost always best addressed with more than a basic symbolic representation. The reader demands to know why the converse does not hold! Examples of situations where the converse does not hold are almost always useful.

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  • $\begingroup$ I fail to understand why an important fact should not be expressed in a single symbol - after all, A <=> B for A => B and B => A is used quite frequently. $\endgroup$ – DaveBall aka user750378 Jul 31 '13 at 18:28
  • $\begingroup$ Reason #1: Symbols can be interpreted different ways under different contexts. Reason #2: Symbols obscure deeper meanings, words enhance them. Symbols should be used as shortcuts when the meaning of something is too trivial, too obvious, or too unimportant to describe with words. One uses symbols when one must repeatedly convey the same concept. But in many cases, repeated concepts are sufficiently interesting to demand words! $\endgroup$ – Emily Jul 31 '13 at 18:51
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    $\begingroup$ Or, in more cynical terms, people tend to bury things in symbols when they don't understand what the symbols represent or what the consequences of the relationship are. If you write things out in words, there is no such doubt, and there is no such ambiguity. $\endgroup$ – Emily Jul 31 '13 at 18:52

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