# Talk:Inverse of Product of Subsets of Group

To prove that $A \subseteq B$ one generally uses an argument of the form: Suppose $x \in A$. Then … $x \in B$. Alternatively one may use proof by contradiction, or some specialized theorem. I see here nothing of this nature.

Furthermore: I see in many cases on this site (including this proof, which has bigger problems) things that read (roughly)

$x \in A \implies x \in B \implies A \subseteq B$.

This does not make sense. $x \in A$ does not imply $A \subseteq B$. What is meant is more in the nature of

$x \in A \implies x \in B$ so $A \subseteq B$.

I'm not familiar with all the logical notation, but perhaps this would be written

$x \in A \implies x \in B \vdash A \subseteq B$. --Dfeuer (talk) 10:00, 4 February 2013 (UTC)
I see what you mean. The conclusion in the second eqn template should be split off. It would require two different interpretations of the column of $\implies$s to let it make sense. --Lord_Farin (talk) 10:44, 4 February 2013 (UTC)
$x \in A \implies x \in B \implies A \subseteq B$.
$y^{-1} \circ x^{-1} \in \left({X \circ Y}\right)^{-1} \implies Y^{-1} \circ X^{-1} \subseteq \left({X \circ Y}\right)^{-1}$? What on earth's wrong with that? The "first, note that" has just demonstrated that $y^{-1} \circ x^{-1} \in Y^{-1} \circ X^{-1}$. Job done. Where's the problem? --prime mover (talk) 13:09, 4 February 2013 (UTC)
What are $x$ and $y$. It's not strictly speaking a valid argument as it stands, sorry. The final $\implies$ depends on all the ones before, which gives it a different meaning from the others. This should be explicated. --Lord_Farin (talk) 13:15, 4 February 2013 (UTC)
$x$ and $y$ are elements of $X$ and $Y$ respectively, as given in the first part. Admittedly, this can be done with explaining with "let $x$ and $y$ be arbitrary elements of $X$ and $Y$ respectively." If there are problems with this proof they are minor and just require a bit of tidying up. --prime mover (talk) 13:32, 4 February 2013 (UTC)