B-Algebra Identity: x (y z) = (x (0 z)) y
Jump to navigation
Jump to search
Theorem
Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
- $\forall x, y, z \in X: x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$
Proof
Let $x, y, z \in X$.
Then:
\(\ds \paren {x \circ \paren {0 \circ z} } \circ y\) | \(=\) | \(\ds x \circ \paren {y \circ \paren {0 \circ \paren {0 \circ z} } }\) | $B$-Algebra Axiom $(\text A 2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {y \circ z}\) | Identity: $y \circ \paren {0 \circ \paren {0 \circ z} } = y \circ z$ |
Hence the result.
$\blacksquare$
Also see
This identity is comparable to $B$-Algebra Axiom $(\text A 3)$.