# Algebra of Sets: Proof of absorption laws without using DeMorgan's laws?

I'm trying to prove set algebras's absoption laws without using DeMorgan's laws.

Absoption laws :

1. $$A \cup (A \cap B) = A$$
2. $$A \cap (A \cup B) = A$$

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

• identity laws

• complement laws

• duality law

• 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)

• etc.

• The "axioms" say almost nothing on complements so it's no wonder you cannot show de Morgan, I think. – Henno Brandsma Apr 17 at 21:03
• @HennoBrandsma. I first would like to prove absoption laws, in order to go from more basic to less basic, as if I were constructing a deductive system. – Eleonore Saint James Apr 17 at 21:07

$$A \cup (A \cap B) \subseteq A$$ follows from $$A \subseteq A$$ and $$A \cap B \subseteq A$$ right away, doesn't it? (interpreting $$\cap$$ as $$\land$$ (the infimum in a lattice) and $$\cup, \lor$$ as the supremum in a lattice )
Trying to use your linked axioms: $$A \cup (A \cap B)$$ by distributivity equals $$(A \cup A) \cap (A \cup B)$$ and by idempotency (proved on your page from your axioms) this equals $$A \cap (A \cup B)$$ which equals by distributivity again : $$(A \cap A) \cup (A \cap B)$$ and we're back at $$A \cup (A \cap B)$$ again. The linked page does not give a definition of inclusion really, but some equivalent statements for $$A \subseteq B$$, that I think do not derive from these axioms but from simple "element considerations".