Below are the important general identities of sets along with clear, step-by-step proofs using element-wise arguments.
These are exam-oriented proofs and easy to write.
⭐ 1. Idempotent Laws
(a) A ∪ A = A
Proof:
Let ( x \in A \cup A ).
By definition of union:
( x \in A ) or ( x \in A ).
In either case, ( x \in A ).
Thus, ( A \cup A \subseteq A ).
Now let ( x \in A ).
Since A is part of ( A \cup A ), ( x \in A \cup A ).
Thus, ( A \subseteq A \cup A ).
Hence,
[
A \cup A = A
]
(b) A ∩ A = A
Proof:
Let ( x \in A \cap A ).
Intersection means ( x \in A ) and ( x \in A ).
So clearly, ( x \in A ).
Thus, ( A \cap A \subseteq A ).
Now take any ( x \in A ).
Then ( x \in A \cap A ).
Thus, ( A \subseteq A \cap A ).
Therefore,
[
A \cap A = A
]
⭐ 2. Commutative Laws
(a) A ∪ B = B ∪ A
Proof:
Let ( x \in A \cup B ).
Then ( x \in A ) or ( x \in B ).
So ( x \in B \cup A ).
Thus, ( A \cup B \subseteq B \cup A ).
Similarly, reverse the argument to show
( B \cup A \subseteq A \cup B ).
Hence,
[
A \cup B = B \cup A
]
(b) A ∩ B = B ∩ A
Proof is similar:
( x \in A \cap B \Rightarrow x \in A \text{ and } x \in B \Rightarrow x \in B \cap A )
Thus,
[
A \cap B = B \cap A
]
⭐ 3. Associative Laws
(a) (A ∪ B) ∪ C = A ∪ (B ∪ C)
Proof:
Let ( x \in (A \cup B) \cup C ).
Then
[
x \in (A \cup B) \quad \text{or} \quad x \in C
]
which means:
[
(x \in A \text{ or } x \in B) \quad \text{or} \quad x \in C
]
Rearranging:
[
x \in A \quad \text{or} \quad (x \in B \text{ or } x \in C)
]
So,
[
x \in A \cup (B \cup C)
]
Reverse direction is similar.
Thus,
[
(A \cup B) \cup C = A \cup (B \cup C)
]
(b) (A ∩ B) ∩ C = A ∩ (B ∩ C)
Same method using “and”.
⭐ 4. Distributive Laws
(a) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Proof:
(i) Show LHS ⊆ RHS
Let ( x \in A \cap (B \cup C) ).
This means:
- ( x \in A )
- ( x \in B \cup C ), so ( x \in B ) or ( x \in C )
Case 1: ( x \in B )
→ ( x \in A \cap B )
Case 2: ( x \in C )
→ ( x \in A \cap C )
Thus, in either case ( x \in (A ∩ B) ∪ (A ∩ C) ).
So LHS ⊆ RHS.
(ii) Show RHS ⊆ LHS
Let ( x \in (A ∩ B) ∪ (A ∩ C) ).
Case 1: ( x \in A ∩ B )
→ x ∈ A and x ∈ B
→ x ∈ B ∪ C
→ x ∈ A ∩ (B ∪ C)
Case 2: ( x ∈ A ∩ C ): similar.
Thus RHS ⊆ LHS.
Hence,
[
A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
]
(b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Proof is similar.
⭐ 5. De Morgan’s Laws
(a) (A ∪ B)’ = A’ ∩ B’
Proof:
Take ( x \in (A \cup B)’ ).
This means
[
x \notin A \cup B
]
→ x ∉ A and x ∉ B
→ x ∈ A’ and x ∈ B’
→ x ∈ A’ ∩ B’
Thus LHS ⊆ RHS.
Reverse:
Let ( x ∈ A’ ∩ B’ ).
Then x ∉ A and x ∉ B
→ x ∉ A ∪ B
→ x ∈ (A ∪ B)’
Thus RHS ⊆ LHS.
Therefore,
[
(A \cup B)’ = A’ \cap B’
]
(b) (A ∩ B)’ = A’ ∪ B’
Proof is similar using “or”.
⭐ 6. Absorption Laws
(a) A ∪ (A ∩ B) = A
Proof:
(i) LHS ⊆ A:
If x ∈ A, done.
If x ∈ A ∩ B, then x ∈ A.
So all elements of LHS are in A.
(ii) A ⊆ LHS:
If x ∈ A, then x ∈ A ∪ (A ∩ B).
Thus,
[
A \cup (A \cap B) = A
]
(b) A ∩ (A ∪ B) = A
Similar reasoning.
⭐ 7. Complement Laws
- ( A \cup A’ = U )
- ( A \cap A’ = \phi )
Proofs follow directly from the definition of complement.