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Suppose we know that there exists widgets. For any widget $X$, we have that the statements $A(X)$ and $B(X)$. Suppose it's true that for any $X$, $A(X) \not \Rightarrow B(X)$. How do I interpret this? I'd take it to mean that if there exists $X$ such that $A(X)$ is true, then it must be for some of these cases that $B(X)$ is false. Or does it mean that if $A(X)$ is true for all $X$, then there exists $X$ such that $B(X)$ is false?

I mean if we replace $A(X)$ by $A$ and $B(X)$ by $B$ and exclude widget specific properties, then if $A \not \Rightarrow B$ and $A$ is true, then I understand that there must exist cases for which $B$ is false. But if we include this widget specifier, then I'm not sure how to interpret this anymore.

If this is ill-posed or ambiguous (which I'm sure it is), feel free to point out what needs to be clarified.

Benjamin Lindqvist
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2 Answers2

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The closest interpretation of what you say is, to me, $\forall X(\neg(A(X) \to B(X)))$. In this case it means that for each widget $W$ individually, you have $\neg(A(W) \to B(W))$. Using the truth table of $\to$, we see that this must mean that $A(W)$ holds, but $B(W)$ does not. Thus, for each widget $A$ holds while $B$ does not.

Mees de Vries
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  • Really? $\forall X, \neg (A(X)\Rightarrow B(X))$ implies that $A(X)$ always holds and $B(X)$ never holds? – Benjamin Lindqvist Sep 29 '16 at 21:14
  • Yes. In fact, $A(X) \to B(X)$ is the same as $\neg A(X) \lor B(X)$, so by De Morgan it holds that $\neg(A(X) \to B(X))$ is the same as $A(X) \land \neg B(X)$. Thus the whole sentence means $\forall X(A(X) \land \neg B(X))$. – Mees de Vries Sep 29 '16 at 21:21
  • You know this whole logic business is messing with me. I am going to reformulate the question as clearly as I can to make sure I'm not messing anything up. – Benjamin Lindqvist Sep 29 '16 at 21:41
  • http://math.stackexchange.com/questions/1947153/lack-of-implication-and-logical-quantifiers. Apologies for the misformulated question. – Benjamin Lindqvist Sep 29 '16 at 21:55
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Without the quantifies and variables to cloud the issue...

$A\Rightarrow B \equiv \neg [A\land \neg B]$

So $A\nRightarrow B \equiv A \land \neg B$

Dan Christensen
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