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Is there a mathematical symbol for 'implies and is not implied by'?

ie.

$$\Rightarrow \land \nLeftarrow$$

Context:

To provide extra emphasis in proofs with lines such as

$$x = 1 \,\,\,\,\,(\Rightarrow \land \nLeftarrow)\,\,\,\,\, x^2 = 1$$

Shuri2060
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2 Answers2

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flawr has created one way to say what you're trying to say, but it isn't any better than your suggestion, in that neither will be readily understood unless you pre-define what the notation means.

To be understood clearly, simply write $$(A \to B) \land \lnot (B\to A)\tag{Use for greatest clarity (1)}$$

No need to create fancy notation for $(1)$, and in doing so, it will likely confuse otheres as to what you're trying to say.

Edit

In response to the helpful comment from @T.Gunn, below:

If we take the question as a universal statement, with predicate "=", we can express the asker's $\Rightarrow$ as follows $$\forall x ((x=1) \to (x^2 = 1))$$ whereas, we have $\not\Leftarrow$ means: $$\lnot \forall x((x^2=1)\to (x=1)) \equiv \exists x ((x^2 =1) \land (x\neq 1))\tag{$\dagger$}$$

$\dagger$: In this case, the existence of $x=-1$ verifies that $$\exists x (x^2 =1) \land (x\neq 1)$$

Given this universal interpretation, the notation used by the asker is not appropriate, as it might be for a propositional implication. What we can say, is $$\forall x ((x=1) \to (x^2 = 1)) \land \lnot \forall x((x^2=1)\to (x=1))$$

amWhy
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    In the case with the two statements $x = 1$ and $x^2 = 1$ we are actually looking at the statements $\forall x (x = 1 \to x^2 = 1)$ and $\neg \forall x (x^2 = 1 \to x = 1)$. I.e. the not is on the other side of a universal quantifier. – Trevor Gunn Jul 10 '17 at 20:57
  • Yeah, there is almost always an implicit quantifier in statements like this which make them not just equivalent to $\neg p\wedge q$. – Eric Wofsey Jul 10 '17 at 20:59
  • @T.Gunn, EricWofsey Well, the same applies to the other answer (propositional A, propositional B), and, like Hagen, (given the OP's failure to claim it was a a universal statement) noted the exact same thing as I (the comment @ +2). So, please attend to both the comment, and the other answer, else I can only conclude that you are are targeting me alone. – amWhy Jul 10 '17 at 21:05
  • @T.Gunn, I've edited my post accordingly, thanks to your comment. – amWhy Jul 10 '17 at 21:33
  • @amWhy I don't think there's anything wrong with flawr's answer. If you read $x^2 = 1$ does not imply $x = 1$ that statement has implicit quantifiers. One can naively write this as $(x^2 = 1) \not\to (x = 1)$ but rigorously this means something different (it has a free variable). I think the take away is "more words, less symbols." – Trevor Gunn Jul 10 '17 at 21:34
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    T. Gunn: My primary take-away was simply to note that neither $A\Rightarrow \land \nLeftarrow B$ nor $A \begin{array}{c} \Rightarrow \ \nLeftarrow \end{array} B$ is as clear as simply stating (1) $(A\Rightarrow B)\land \lnot(B\rightarrow A)$. Or (2) simply stating $(x=1)$ implies $(x^2 = 1)$, but it's not the case that $x^2 =1$ implies $x = 1 $ – amWhy Jul 10 '17 at 21:46
  • @amWhy, T.Gunn Thanks for your answer/comments and making that clarification - yes, I wondered myself when I asked that question as it didn't quite sound right. Indeed, the qualifiers are necessary for absolute clarity which I hadn't thought of at the time – Shuri2060 Jul 10 '17 at 22:26
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I don't think there is an own symbol, but to express $A \implies B$ and $B \,\,\,\,\,\,\not\!\!\!\!\!\implies A$ I've seen

$\renewcommand{\arraystretch}{0.1}$

$$A \begin{array}{c} \Rightarrow \\ \nLeftarrow \end{array} B$$

flawr
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