Every Operation is Distributive over Right Operation
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Theorem
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\to$ denote the right operation on $S$.
Then $\circ$ is distributive over $\to$.
Proof
\(\ds \forall a, b, c \in S: \, \) | \(\ds a \circ \paren {b \to c}\) | \(=\) | \(\ds a \circ c\) | Definition of Right Operation | ||||||||||
\(\ds \forall a, b, c \in S: \, \) | \(\ds \paren {a \circ b} \to \paren {a \circ c}\) | \(=\) | \(\ds a \circ c\) | Definition of Right Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall a, b, c \in S: \, \) | \(\ds a \circ \paren {b \to c}\) | \(=\) | \(\ds \paren {a \circ b} \to \paren {a \circ c}\) | from above |
\(\ds \forall a, b, c \in S: \, \) | \(\ds \paren {a \to b} \circ c\) | \(=\) | \(\ds b \circ c\) | Definition of Right Operation | ||||||||||
\(\ds \forall a, b, c \in S: \, \) | \(\ds \paren {a \circ c} \to \paren {b \circ c}\) | \(=\) | \(\ds b \circ c\) | Definition of Right Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall a, b, c \in S: \, \) | \(\ds \paren {a \to b} \circ c\) | \(=\) | \(\ds \paren {a \circ c} \to \paren {b \circ c}\) | from above |
Hence the result by definition of distributive.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.26 \ \text {(a)}$