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If I have two statement, say $A$ and $B$. From which, I formed two formulae:

$F_1$: ~$A$ $\&$ ~$B$

$F_2$: ~$A$ $\mid$ ~$B$

Do $F_1$ and $F_2$ entail each other?

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    If $|$ is the [Sheffer stroke](https://en.wikipedia.org/wiki/Sheffer_stroke), than $\lnot A \mid \lnot B$ is simply : $A \land B$. Thus : $F_2 \nvDash F_1$, because otherwise $A \land B \vDash \lnot A \land \lnot B$, which is not. – 2017-01-27

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I assume that your $|$ is meant for 'or' (in programming it often is), and an inclusive or at that (again, in programming it often is used as such).

So, using more 'standard' logical symbols, the statements are:

$F_1: \neg A \land \neg B$

$F_2: \neg A \lor \neg B$

Now, a statement $\phi$ entails a statement $\psi$ if and only if $\psi$ has to be true if we assume $\phi$ is true.

So in this case:

$F_1$ entails $F_2$, because if we assume that $F_1$ is true, then $\neg A \land \neg B$ is true, hence in particular $\neg A$, and that means that $\neg A \lor \neg B$ is true (think about it: if $P$ is true, then of course $P \lor Q$ will be true as well)

On the other hand, $F_2$ does not entail $F_1$, since we could have a situation where $A$ is true, and $B$ is false: that would mean that $\neg B$ is true, and hence $\neg A \lor \neg B$ (i.e. $F_2$) is true, but since $A$ is true, clearly $\neg A \land \neg B$ (i.e. $F_1$) is false. So this shows that it is possible for $F_2$ to be true but $F_1$ to be false, so $F_2$ does not entail $F_1$.

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    Very informative. I had the second confusion. Thank you sir. – 2017-01-28