De Morgans Law of Set Theory Proof - Math Theorems Statement: Demorgan's First Law: (A B)' = (A)' ∩ (B)'
The first law states that the complement of the union of two sets is the intersection of the complements. Proof : (A ∪ B)' = (A)' ∩ (B)' Consider x ∈ (A ∪ B)' If x ∈ (A ∪ B)' then x ∉ A ∪ B (x ∈ A ∪ B)' (x ∈ A ∪ x ∈ B)' (x ∈ A)' ∩ (x ∈ B)' (x ∉ A) ∩ (x ∉ B) (x ∈ A') ∩ (x ∈ B') x ∈ A' ∩ B' Therefore, (A B)' = (A)' ∩ (B)'
Definition of compliment Definition ∉ Definition of ∪ Definition of ∉ Definition of compliment Definition of ∩
Demorgan's Second Law: (A ∩ B)' = (A)' (B)'
The second law states that the complement of the intersection of two sets is the un ion of the complements. Proof : (A ∩ B)' = (A)' ∪ (B)' Consider x ∈ (A ∩ B)' If x ∈ (A ∩ B)' then x ∉ A ∩ B (x ∈ A ∩ B)' (x ∈ A ∩ x ∈ B)' (x ∈ A)' ∪ (x ∈ B)' (x ∉ A) ∪ (x ∉ B) (x ∈ A') ∪ (x ∈ B') x ∈ A' ∪ B' Therefore, (A ∩ B)' = (A)' (B)'
Definition of compliment Definition of ∉ Definition of ∩ Definition of ∉ Definition of compliment Definition of ∪
Algebra Commutative Property Statement: First Law : First law states that the union of two sets is the same n o matter what the order is in the equation. A B=B A
Proof : A B = B A Consider the first law, A ∪ B = B ∪ A Let x ∈ A ∪ B. If x ∈ A ∪ B then x ∈ A or x ∈ B x ∈ A or x ∈ B x ∈ B or x ∈ A [according to definition of union] x ∈ B ∪ A x ∈ A ∪ B => x ∈ B ∪ A Therefore, A ∪ B ⊂ B ∪ A --- 1
Consider the first law in reverse, B ∪ A = A ∪ B Let x ∈ B ∪ A. If x ∈ B ∪ A then x ∈ B or x ∈ A x ∈ B or x ∈ A x ∈ A or x ∈ B [according to definition of union] x ∈ A ∪ B x ∈ B ∪ A => x ∈ A ∪ B Therefore, B ∪ A ⊂ A ∪ B --- 2 From equation 1 and 2 we can prove A B=B A Second Law : Second law states that the intersection of two sets is the same no matter what the order is in the equation. A∩B=B∩A Proof : A ∩ B = B ∩ A Consider the second law, A ∩ B = B ∩ A Let x ∈ A ∩ B. If x ∈ A ∩ B then x ∈ A and x ∈ B x ∈ A and x ∈ B x ∈ B and x ∈ A [according to definition of intersection] x∈B∩A x ∈ A ∩ B => x ∈ B ∩ A Therefore, A ∩ B ⊂ B ∩ A --- 3
Consider the second law in reverse, B ∩ A = A ∩ B Let x ∈ B ∩ A. If x ∈ B ∩ A then x ∈ B and x ∈ A x ∈ B and x ∈ A x ∈ A and x ∈ B [according to definition of intersection] x∈A∩B
x ∈ B ∩ A => x ∈ A ∩ B Therefore, B ∩ A ⊂ A ∩ B --- 4 From equation 3 and 4 we can prove the Commutative Property A∩B=B∩A
Associative Law of Set Theory Proof Statement: First Law: First law states that the intersection of a set to the intersection of two other sets is the same. (A ∩ B) ∩ C = A ∩ (B ∩ C) Proof : In the first law (A ∩ B) ∩ C = A ∩ (B ∩ C) Step 1: Let us take the L.H.S, (A ∩ B) ∩ C Let x ∈ (A ∩ B) ∩ C. If x ∈ (A ∩ B) ∩ C then x ∈ (A and B) and x ∈ C x ∈ (A and B) and x ∈ C x ∈ (A and B) implies x ∈ A and x ∈ B So, we have x ∈ A, x ∈ B and x ∈ C x ∈ A and x ∈ (B and C) x ∈ A and (B and C) x ∈ A ∩ (B ∩ C) x ∈ (A ∩ B) ∩ C => x ∈ A ∩ (B ∩ C) (A ∩ B) ∩ C A ∩ (B ∩ C)--- 1 Step 2: Let us take the R.H.S, (A ∩ B) ∩ C Let x ∈ A ∩ (B ∩ C). If x ∈ A ∩ (B ∩ C) then x ∈ A and x ∈ (B and C) x ∈ A and x ∈ (B and C) x ∈ (B and C) implies x ∈ B and x ∈ C So, we have x ∈ A, x ∈ B and x ∈ C x ∈ (A and B) and x ∈ C x ∈ (A and B) and C x ∈ (A ∩ B) ∩ C x ∈ A ∩ (B ∩ C) => x ∈ (A ∩ B) ∩ C A ∩ (B ∩ C) (A ∩ B) ∩ C--- 2
From equation 1 and 2 (A ∩ B) ∩ C = A ∩ (B ∩ C) Hence, associative law of sets for intersection has been p roved. Second Law:
Second law states that the union of a set to the union of two other sets is the same. (A B) C = A (B C) Proof : In the second law (A ∪ B) ∪ C = A ∪ (B ∪ C) Step 1: Let us take the L.H.S, (A ∪ B) ∪ C Let x ∈ (A ∪ B) ∪ C. If x ∈ (A ∪ B) ∪ C then x ∈ (A or B) or x ∈ C x ∈ (A or B) or x ∈ C x ∈ (A or B) implies x ∈ A or x ∈ B So, we have x ∈ A or x ∈ B or x ∈ C x ∈ A or x ∈ (B or C) x ∈ A or (B or C) x ∈ A ∪ (B ∪ C) x ∈ (A ∪ B) ∪ C => x ∈ A ∪ (B ∪ C) (A B) C A (B C)--- 3 Step 2: Let us take the R.H.S, (A ∪ B) ∪ C Let x ∈ A ∪ (B ∪ C). If x ∈ A ∪ (B ∪ C) then x ∈ A or x ∈ (B or C) x ∈ A or x ∈ (B or C) x ∈ (B or C) implies x ∈ B or x ∈ C So, we have x ∈ A or x ∈ B or x ∈ C x ∈ (A or B) or x ∈ C x ∈ (A or B) or C x ∈ (A ∪ B) ∪ C x ∈ A ∪ (B ∪ C) => x ∈ (A ∪ B) ∪ C A (B C) (A B) C--- 4
From equation 3 and 4 (A B) C = A (B C) Hence, associative law of sets for union has been proved.
Distributive Law Property of Set Theory Proof First Law : A (B ∩ C) = (A
B) ∩ (A
C)
First law states that taking the union of a set to the intersection of two other sets is the same as tak ing the union of the original set and both the other two sets separately, and then taking the intersection of the results. Proof : A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Let x ∈ A ∪ (B ∩ C). If x ∈ A ∪ (B ∩ C) then x is either in A or in (B and C).
x ∈ A or x ∈ (B and C) x ∈ A or {x ∈ B and x ∈ C} {x ∈ A or x ∈ B} and {x ∈ A or x ∈ C} x ∈ (A or B) and x ∈ (A or C) x ∈ (A ∪ B) ∩ x ∈ (A ∩ C) x ∈ (A ∪ B) ∩ (A ∪ C) x ∈ A ∪ (B ∩ C) => x ∈ (A ∪ B) ∩ (A ∪ C) Therefore, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C)--- 1 Let x ∈ (A ∪ B) ∩ (A ∪ C). If x ∈ (A ∪ B) ∩ (A ∪ C) then x is in (A or B) and x is in (A or C). x ∈ (A or B) and x ∈ (A or C) {x ∈ A or x ∈ B} and {x ∈ A or x ∈ C} x ∈ A or {x ∈ B and x ∈ C} x ∈ A or {x ∈ (B and C)} x ∈ A ∪ {x ∈ (B ∩ C)} x ∈ A ∪ (B ∩ C) x ∈ (A ∪ B) ∩ (A ∪ C) => x ∈ A ∪ (B ∩ C) Therefore, (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C)--- 2 From equation 1 and 2 A (B ∩ C) = (A B) ∩ (A Second Law : A ∩ (B C) = (A ∩ B)
C)
(A ∩ C)
Second law states that taking the intersection of a set to the union of two other sets is the same as taking the intersection of the original set and both the other two sets separately, and then taking the union of the results. Proof : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) Let x ∈ A ∩ (B ∪ C). If x ∈ A ∩ (B ∪ C) then x ∈ A and x ∈ (B or C). x ∈ A and {x ∈ B or x ∈ C} {x ∈ A and x ∈ B} or {x ∈ A and x ∈ C} x ∈ (A ∩ B) or x ∈ (A ∩ C) x ∈ (A ∩ B) ∪ (A ∩ C) x ∈ A ∩ (B ∪ C) => x ∈ (A ∩B) ∪ (A ∩ C)
Therefore, A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C)--- 3 Let x ∈ (A ∩ B) ∪ (A ∩ C). If x ∈ (A ∩ B) ∪ (A ∩ C) then x ∈ (A ∩ B) or x ∈ (A ∩ C).
x ∈ (A and B) or (A and C) {x ∈ A and x ∈ B} or {x ∈ A and x ∈ C} x ∈ A and {x ∈ B or x ∈ C} x ∈ A and x ∈ (B or C) x ∈ A ∩ (B ∪ C) x ∈ (A ∩ B) ∪ (A ∩ C) => x ∈ A ∩ (B ∪ C) Therefore, (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C)--- 4 From equation 3 and 4 A ∩ (B C) = (A ∩ B)
(A ∩ C)
Hence, distributive law property of sets theory has been proved. thank you for your reply so would the whole proof be 1.A ∩ (A ∪ B) is a subset of A x is a element in A ∩ (A ∪ B) x is a element in A by definition of intersection Therefore A ∩ (A ∪ B) is a subset of A 2.A is a subset of A ∩ (A ∪ B) x is a element in A x is a element in A ∩ (A ∪ B) by definition of intersection Therefore A is a subset of A ∩ (A ∪ B) 3.Since A ∩ (A ∪ B) is a subset of A and A is a subset of A ∩ (A ∪ B), then A ∩ (A ∪ B) = A
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ABSOTPTION
Theorem S ∩(S ∪T )=S
Proof 1 ⟹
S ⊆(S ∪T )
Set is Subset of Union
S ∩(S ∪T )=S
Intersection with Subset is Subset
■
Proof 2 x∈S ∩(S ∪T )
⟺
x∈S ∧( x∈S ∨ x∈T )
⟺
x∈S
Definitions of Intersection and Union
