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 (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).Subproof that (A ∪ B) ∩ C ⊆ (A ∩ C) ∪ (B ∩ C):
Suppose x ∈ (A ∪ B) ∩ C.
Then, by definition of intersection, x ∈ (A ∪ B) ∧ x ∈ C.
Then, x ∈ (A ∪ B) says x ∈ A ∨ x ∈ B. So, consider these as two cases:
In either case, x ∈ (A ∩ C) ∪ (B ∩ C). So, (A ∪ B) ∩ C ⊆ (A ∩ C) ∪ (B ∩ C).Case x ∈ A:
Then, applying definition of "and" gives that x ∈ A ∧ x ∈ C.
So, by definition of intersection, x ∈ A ∩ C.
Next, by ∨ Introduction, we can say x ∈ A ∩ C ∨ x ∈ B ∩ C. But this meets the definition of union, so we may conclude x ∈ (A ∩ C) ∪ (B ∩ C).Case x ∈ B:
Then, applying definition of "and" gives that x ∈ B ∧ x ∈ C.
So, by definition of intersection, x ∈ B ∩ C.
Next, by ∨ Introduction, we can say x ∈ A ∩ C ∨ x ∈ B ∩ C. But this meets the definition of union, so we may conclude x ∈ (A ∩ C) ∪ (B ∩ C).
Subproof that (A ∩ C) ∪ (B ∩ C) ⊆ (A ∪ B) ∩ C:
Suppose x ∈ (A ∩ C) ∪ (B ∩ C).
Then, by definition of union, x ∈ (A ∩ C) ∨ x ∈ (B ∩ C). So, we'll consider two cases:
In either case, x ∈ (A ∪ B) ∩ C. So, (A ∩ C) ∪ (B ∩ C) ⊆ (A ∪ B) ∩ C.Case x ∈ (A ∩ C):
Then, by definition of intersection, x ∈ A ∧ x ∈ C.
In particular (via ∧ Elimination), x ∈ A. We can apply ∨ Introduction to claim x ∈ A ∨ x ∈ B. So, by definition of union, x ∈ A ∪ B.
Also, we have that x ∈ C. Combining this with the last result gives x ∈ A ∪ B ∧ x ∈ C. We can apply the definition of intersection to get x ∈ (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 claim x ∈ A ∨ x ∈ B. So, by definition of union, x ∈ A ∪ B.
Also, we have that x ∈ C. Combining this with the last result gives x ∈ A ∪ B ∧ x ∈ C. We can apply the definition of intersection to get x ∈ (A ∪ B) ∩ C.
