I'm trying to prove set algebras's absoption laws without using DeMorgan's laws.
Absoption laws :
- $A \cup (A \cap B) = A$
- $A \cap (A \cup B) = A$
Is this possible?
I would like to prove these laws using only set algebra's laws , without analysing the sentences set theoretically in terms of membership relation.
I can prove , for example, that $A \cup ( A \cap B)$ is included in $A$ , by proving that
$( A \cup (A \cap B)) \cap A^\complement = \emptyset$
( here I use the fact that : $X \subseteq Y \iff X \cap Y^\complement= \emptyset$).
But the reverse inclusion, applying the same strategy, would require me to use De Morgan's laws).
Following Wikipedia I take
(1) As axioms & definitions
commutativity of U and Inter
associativity of U and Inter
distributivity of U over Inter , and reciprocally
definition of inclusion
definition of equality ( as reciprocal inclusion)
(2) As derived laws ( provable using axioms and definitions)
idempotent laws for U and Inter ( Which I managed to prove)
domination laws ( Which I managed to prove)
absoption laws ( Here, I am stuck)
DeMorgan's laws ( Here, also stuck)