One-to-Many Image of Set Difference/Corollary 2
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Theorem
Let $\RR \subseteq S \times T$ be a relation which is one-to-many.
Let $A$ be a subset of $S$.
Then:
- $\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
where $\complement$ (in this context) denotes relative complement.
In the language of direct image mappings this can be presented as:
- $\forall A \in \powerset S: \map {\paren {\complement_{\Img \RR} \circ \RR^\to} } A = \map {\paren {\RR^\to \circ \complement_S} } A$
Proof
By definition of the image of $\RR$:
- $\Img \RR = \RR \sqbrk S$
So, when $B = S$ in One-to-Many Image of Set Difference: Corollary 1:
- $\relcomp {\Img \RR} {\RR \sqbrk A} = \relcomp {\RR \sqbrk S} {\RR \sqbrk A}$
Hence:
- $\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
means exactly the same thing as:
- $\relcomp {\RR \sqbrk S} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
that is:
- $\RR \sqbrk S \setminus \RR \sqbrk A = \RR \sqbrk {S \setminus A}$
Hence the result from One-to-Many Image of Set Difference.
$\blacksquare$