Could someone check this proof? If c\b subset c\a, then prove a subset b

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Assume c\b is a subset of c\a. This means if x Є c Λ (Not Є) b, then it is an Є c Λ (Not Є) a.

Assume x Є c Λ (Not Є) b, but is Not Є c Λ (Not Є) a. Then x Є c Λ a. But this contradicts,
c\b is a subset of c\a. Therefore, a must be subset of b.
 
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It doesn't look right to me. Why would you negate the hypothesis ? This is not reductio ad absurdum.

Let's negate the conclusion: a\not\subset b, which means that \exists x\in a, so that x\not\in b. But by hypothesis, \forall x\not\in b, x\not\in a. Contradiction, right ?
 
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You are correct. I see the difference. Thank you for the help.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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