distributive law states that, For all sets

*A*,

*B*, and

*C*,

**Distributive law:-****AU (B∩C) = (AUB) ∩ (AUC)**

**A∩ (BUC) = (A∩B) U (A∩C)**

**Proof**

As we have (## Subproof that (

A∪B) ∩C⊆ (A∩C) ∪ (B∩C):

Supposex∈ (A∪B) ∩C.

Then, by definition of intersection,x∈ (A∪B) ∧x∈C.

Then,x∈ (A∪B) saysx∈A∨x∈B. So, consider these as two cases:

In either case,## Case

x∈A:Then, applying definition of "and" gives thatx∈A∧x∈C.

So, by definition of intersection,x∈A∩C.

Next, by ∨ Introduction, we can sayx∈A∩C∨x∈B∩C. But this meets the definition of union, so we may concludex∈ (A∩C) ∪ (B∩C).## Case

x∈B:Then, applying definition of "and" gives thatx∈B∧x∈C.

So, by definition of intersection,x∈B∩C.

Next, by ∨ Introduction, we can sayx∈A∩C∨x∈B∩C. But this meets the definition of union, so we may concludex∈ (A∩C) ∪ (B∩C).x∈ (A∩C) ∪ (B∩C). So, (A∪B) ∩C⊆ (A∩C) ∪ (B∩C).

## Subproof that (

A∩C) ∪ (B∩C) ⊆ (A∪B) ∩C:

Supposex∈ (A∩C) ∪ (B∩C).

Then, by definition of union,x∈ (A∩C) ∨x∈ (B∩C). So, we'll consider two cases:

In either case,## Case

x∈ (A∩C):Then, by definition of intersection,x∈A∧x∈C.

In particular (via ∧ Elimination),x∈A. We can apply ∨ Introduction to claimx∈A∨x∈B.So, by definition of union,x∈A∪B.

Also, we have thatx∈C. Combining this with the last result givesx∈A∪B∧x∈C. We can apply the definition of intersection to getx∈ (A∪B) ∩C.## Case

x∈ (B∩C):Then, by definition of intersection,x∈B∧x∈C.

In particular (via ∧ Elimination),x∈B. We can apply ∨ Introduction to claimx∈A∨x∈B.So, by definition of union,x∈A∪B.

Also, we have thatx∈C. Combining this with the last result givesx∈A∪B∧x∈C. We can apply the definition of intersection to getx∈ (A∪B) ∩C.x∈ (A∪B) ∩C. So, (A∩C) ∪ (B∩C) ⊆ (A∪B) ∩C.

*A*∪

*B*) ∩

*C*⊆ (

*A*∩

*C*) ∪ (

*B*∩

*C*) and (

*A*∩

*C*) ∪ (

*B*∩

*C*) ⊆ (

*A*∪

*B*) ∩

*C*we may conclude (

*A*∪

*B*) ∩

*C*= (

*A*∩

*C*) ∪ (

*B*∩

*C*).

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