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DeMorgan’s Laws
Transformational Rules for 2 Sets
1. Complement of the Union Equals the Intersection of the Complements
a. not (A or B) = not A and not B
2. Complement of the Intersection Equals the Union of the Complements
a. not (A and B) = not A or not B
1
DeMorgan’s Laws Visual
2
Set Notation Refresher
Take 2 Sets A and B
Union = A U B ← Everything in A or B
Intersection = A ∩ B ← Everything in A and B
U = Universal Set (All possible elements in your defined universe)
Complement = A’ Everything not in A, but in the Universal Set
3
Prove DeMorgan’s Law #1
Complement of the Union Equals the Intersection of the Complements
Let P = (A U B)' and Q = A' ∩ B'
Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)'
⇒ x ∉ (A U B)
⇒ x ∉ A and x ∉ B
⇒ x ∈ A' and x ∈ B'
⇒ x ∈ A' ∩ B'
⇒ x ∈ Q
4
Test Problem for DeMorgan’s Law #1
Complement of the Union Equals the Intersection of the Complements
Take 2 sets A and B. (A U B)’ = A’ ∩ B’
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 2, 3, 4, 5}, B = {2, 4, 6}
A U B = {1, 2, 3, 4, 5, 6}, (A U B)’ = {7, 8, 9, 10}
A’ = {6, 7, 8, 9, 10}, B’ = {1, 3, 5, 7, 8, 9, 10}
A’ ∩ B’ = {7, 8, 9, 10}
5
Prove DeMorgan’s Law #2
Complement of the Intersection Equals the Union of the Complements
Let P = (A ∩ B)' and Q = A' U B'
Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A ∩ B)'
⇒ x ∉ (A ∩ B)
⇒ x ∉ A or x ∉ B
⇒ x ∈ A' or x ∈ B'
⇒ x ∈ A' U B'
⇒ x ∈ Q
6
Test Problem for DeMorgan’s Law #2
Complement of the Intersection Equals the Union of the Complements
Take 2 sets A and B. (A ∩ B)’ = A’ U B’
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 2, 3, 4, 5}, B = {2, 4, 6}
A ∩ B = {2, 4}, (A ∩ B)’ = {1, 3, 5, 6, 7, 8, 9, 10}
A’ = {6, 7, 8, 9, 10}, B’ = {1, 3, 5, 7, 8, 9, 10}
A’ U B’ = {1, 3, 5, 6, 7, 8, 9, 10}
7

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How to Prove and Apply De Morgan's Laws

  • 1. DeMorgan’s Laws Transformational Rules for 2 Sets 1. Complement of the Union Equals the Intersection of the Complements a. not (A or B) = not A and not B 2. Complement of the Intersection Equals the Union of the Complements a. not (A and B) = not A or not B 1
  • 3. Set Notation Refresher Take 2 Sets A and B Union = A U B ← Everything in A or B Intersection = A ∩ B ← Everything in A and B U = Universal Set (All possible elements in your defined universe) Complement = A’ Everything not in A, but in the Universal Set 3
  • 4. Prove DeMorgan’s Law #1 Complement of the Union Equals the Intersection of the Complements Let P = (A U B)' and Q = A' ∩ B' Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)' ⇒ x ∉ (A U B) ⇒ x ∉ A and x ∉ B ⇒ x ∈ A' and x ∈ B' ⇒ x ∈ A' ∩ B' ⇒ x ∈ Q 4
  • 5. Test Problem for DeMorgan’s Law #1 Complement of the Union Equals the Intersection of the Complements Take 2 sets A and B. (A U B)’ = A’ ∩ B’ U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {1, 2, 3, 4, 5}, B = {2, 4, 6} A U B = {1, 2, 3, 4, 5, 6}, (A U B)’ = {7, 8, 9, 10} A’ = {6, 7, 8, 9, 10}, B’ = {1, 3, 5, 7, 8, 9, 10} A’ ∩ B’ = {7, 8, 9, 10} 5
  • 6. Prove DeMorgan’s Law #2 Complement of the Intersection Equals the Union of the Complements Let P = (A ∩ B)' and Q = A' U B' Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A ∩ B)' ⇒ x ∉ (A ∩ B) ⇒ x ∉ A or x ∉ B ⇒ x ∈ A' or x ∈ B' ⇒ x ∈ A' U B' ⇒ x ∈ Q 6
  • 7. Test Problem for DeMorgan’s Law #2 Complement of the Intersection Equals the Union of the Complements Take 2 sets A and B. (A ∩ B)’ = A’ U B’ U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {1, 2, 3, 4, 5}, B = {2, 4, 6} A ∩ B = {2, 4}, (A ∩ B)’ = {1, 3, 5, 6, 7, 8, 9, 10} A’ = {6, 7, 8, 9, 10}, B’ = {1, 3, 5, 7, 8, 9, 10} A’ U B’ = {1, 3, 5, 6, 7, 8, 9, 10} 7

Editor's Notes

  1. Read this as the complement of A Union B equals the Intersection of A Complement and B Complement
  2. Read this as the complement of A Union B equals the Intersection of A Complement and B Complement
  3. Read this as the complement of A Union B equals the Intersection of A Complement and B Complement
  4. Read this as the complement of A Union B equals the Intersection of A Complement and B Complement