I am attaching an image, I will award full points to whoever can give me a detai

Question

I am attaching an image, I will award full points to whoever can give me a detailed proof that is easy to understand. I have seen a proof but I am having a hard time understanding. I need this in proof format. Thank you. I am needing you to prove  Theorem 1.36 (Demorgans laws)

Union x epsilon (0,1] [0, x) = p0, 1) and intersection x epsilon (0, 1] [0, x) = {0}. The following important, often used result shows that there is an easy way to get from unions to intersections, and vice versa. Let X be a set and {Ealpha)alpha epsilon A be a collection of subsets of X. If for each E subset X the symbol Ec represents the set XE, then (Union alpha epsilon A Ealpha)c = intersection alpha epsilon A and (intersection alpha epsilon A Ealpha)c = Union alpha epsilon A .

Explanation / Answer

1) Let x in complement of union of E_a , a in A

Then x does not belong to union of E_a, a in A

i.e x does not belong to any of E_a for all a in A

Thus, x belongs to complement of E_a for all a in A

And hence, x belongs to intersection of complement of E_a, a in A

And, we move backward to prove converse part

2) Suppose x belongs to complement of intersection of E_a, a in A

then x does not belong to intersection of E_a, a in A

so, x does not belong to atleast one E_a for some a in A

thus x belongs to the complement of E_a for that a in A

Hence x belongs to union of complements of E_a, a in A

We move backward to prove the converse

Hence proved

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