Subset Relation is Compatible with Subset Product/Corollary 2
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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.
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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.
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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.
$\blacksquare$