How can the complementation law in Table 1 be proven for \stackrel{=}{A} = A?

Bashyboy
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Homework Statement


Prove the complementation law in Table 1 by showing
that \stackrel{=}{A} = A


Homework Equations





The Attempt at a Solution



Well, first I assumed that x is an element of A, so that A = (x | x\in A)

by taking the complement, I got (x | \neg(x\in A) \rightarrow (x | x\notin A)

then, taking the complement of the complement is where I get stuck:

(x | \neg(x \in \overline{A})
 
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I have another one:
Prove the domination laws in Table 1 by showing that
A ∪ U = U

A∪U = {x| x∈A∨x∈U} = {x| x∈ A ∨ T} = {x| T}=U

This is from the solution manual. I understand all but the last step. To me, the last step seems meaningless; how could you infer anything from it?
 
In my original post, the arrow should actually be an equal sign.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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