Math, asked by ahah77, 2 months ago

Prove
Can you guys prove that A subset of C with using Au(BnC)=(AuB)nC?
Thank you :)

Answers

Answered by YourBoyJeff
1

Answer:

For any three sets A, B and C, we are given to prove the following :

A - (B ∩ C) = (A - B) U (A - C).

We know that

any two sets P and Q are equal if and only if both are subsets of each other, that is

P ⊂ Q and Q ⊂ P.

Let us consider that

  x ∈ A - (B ∩ C)

⇒x∈A, x ∉ (B ∩ C)

⇒x∈A, (x∉B or x∉C)

⇒(x∈A, x∉B) or (x∈A, x∉C)

⇒x∈(A-B) or x∈(A-C)

⇒x∈(A-B) ∪ (A-C)

So, A - (B n C) ⊂ (A-B) U (A-C).

Again, let

 x∈(A-B) ∪ (A-C)

⇒x∈(A-B) or x∈(A-C)

⇒(x∈A, x∉B) or (x∈A, x∉C)

⇒x∈A, (x∉B or x∉C)

⇒x∈A, x ∉ (B ∩ C)

⇒x ∈ A - (B ∩ C).

So, A - (B ∩ C) ⊂ (A-B) ∪ (A-C).

Therefore, we get

A - (B n C) = (A - B) U (A - C).

Answered by vikashpatnaik2009
0

Answer:

This chapter reviews certain logical notions and considers the Formulas—that is, expressions that may contain (free) variables and whose logical values depend on values given to these variables. The chapter describes a class K of formulas, called the class of pure set-theoretical formulas. The chapter explains that formulas that have no free variables are called sentences. A sentence from a formula can be obtained by inserting in the front of the formula enough quantifiers to bind all the free variables of the formula. The chapter also introduces one additional primitive notion that, although not strictly necessary, simplifies many statements of set theory. The chapter also explains that in the case of the propositional calculus, there are many formulas involving quantifiers that are true independent of the meaning of atomic formulas out of which they are constructed and independent of the values given to the free variables. Such formulas are called theorems of the predicate

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