Sure, let's prove this using the laws of set theory.

Step 1: Let's take an element x from (A ∪ B)′. By definition of set complement, x is not in A ∪ B.

Step 2: By definition of set union, this means that x is not in A and x is not in B.

Step 3: By definition of set complement again, this means that x is in A′ and x is in B′.

Step 4: By definition of set intersection, this means that x is in A′ ∩ B′.

So, we have shown that if x is in (A ∪ B)′, then x is in A

Question

Sure, let's prove this using the laws of set theory.

Step 1: Let's take an element x from (A ∪ B)′. By definition of set complement, x is not in A ∪ B.

Step 2: By definition of set union, this means that x is not in A and x is not in B.

Step 3: By definition of set complement again, this means that x is in A′ and x is in B′.

Step 4: By definition of set intersection, this means that x is in A′ ∩ B′.

So, we have shown that if x is in (A ∪ B)′, then x is in A

Knowee AI · Accepted Answer

Sure, let's prove this using the laws of set theory.

Step 1: Let's take an element x from (A ∪ B)′. By definition of set complement, x is not in A ∪ B.

Step 2: By definition of set union, this means that x is not in A and x is not in B.

Step 3: By definition of set complement again, this means that x is in A′ and x is in B′.

Step 4: By definition of set intersection, this means that x is in A′ ∩ B′.

So, we have shown that if x is in (A ∪ B)′, then x is in A

Prove that: (A ∪ B)′ ≡ A′ ∩ B′.