Talk:Inverse of Product of Subsets of Group

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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)
Where in this page do we have an instance of:
$x \in A \implies x \in B \implies A \subseteq B$.
? --prime mover (talk) 11:36, 4 February 2013 (UTC)

The third and fourth lines of the second group of implications. But really, that's not even the most confusing thing here. The most confusing thing here is that I don't see where the implication derived in the "first, note that" group is actually used. --Dfeuer (talk) 11:49, 4 February 2013 (UTC)

$y^{-1} \circ x^{-1} \in \left({X \circ Y}\right)^{-1} \implies Y^{-1} \circ X^{-1} \subseteq \left({X \circ Y}\right)^{-1}$? 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)