Structure Induced by Absorbing Operations is Absorbing
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Theorem
Let $\struct {T, \circ, *}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus, \otimes}$ be the structure on $T^S$ induced by $\circ$ and $*$.
Let $\circ$ and $*$ satisfy the absorption law:
- $\forall a, b \in S: a \circ \paren {a * b} = a$
Then the pointwise operations $\oplus$ and $\otimes$ on $T^S$ also satisfy the absorption law:
- $\forall f, g \in T^S: f \oplus \paren {f \otimes g} = f$
Proof
Let $f, g \in T^S$.
Then:
\(\ds \forall x \in S: \, \) | \(\ds \map {\paren {f \oplus \paren {f \otimes g} } } x\) | \(=\) | \(\ds \map f x \circ \map {\paren {f \otimes g} } x\) | Definition of Pointwise Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds \map f x \circ \paren {\map f x * \map g x}\) | Definition of Pointwise Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f x\) | $\circ$ absorbs $*$ |
From Equality of Mappings:
- $f \oplus \paren {f \otimes g} = f$
Since $f, g$ were arbitrary:
- $\forall f, g \in T^S : f \oplus \paren {f \otimes g} = f$
Hence $\oplus$ absorbs $\otimes$ by definition.
$\blacksquare$