Subset Relation is Compatible with Subset Product/Corollary 2
Jump to navigation
Jump to search
Although this article appears correct, it's inelegant. There has to be a better way of doing it.
No need to use power set. $A \subseteq B \subseteq S$ sufficient.
You can help Proof Wiki by redesigning it.
Theorem
Let $\left({S,\circ}\right)$ be a magma.
Let $A,B \in \mathcal P \left({S}\right)$, the power set of $S$.
No need to use power set. $A \subseteq B \subseteq S$ sufficient.
You can help Proof Wiki by redesigning it.
To discuss this proof in more detail, feel free to use the talk page.
If you are able to fix this proof, then when you have done so you can remove this instance of {{Improve}} from the code.
If you would welcome a second opinion as to whether your improvement is correct, add an invitation to {{Proofread}} the page (see the proofread template for usage).
Let $A \subseteq B$.
Let $x \in S$.
Then:
- $x \circ A \subseteq x \circ B$
- $A \circ x \subseteq B \circ x$
Proof
This follows from Subset Relation is Compatible with Subset Product and the definition of the subset product with a singleton.
$\blacksquare$
