Understanding Logical Equivalencies in ~p ^ q and F: Explained Simply

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Everything in here makes sense up and till the last part, where they substituted (I think) F in for ~p ^ q, and then just ignored it and made it equivalent to the last part? Hope I'm not being too vague here.
 
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They observed that ((not p) and p) is always false (F). Then they used that (false or anything)=anything. Is it the 'F' that throwing you off?
 
Oh ok I see so false and anything is always anything. That makes sense. So the F represents a substitution of the previous part but is also used to symbolize false, correct?
 
mohabitar said:
Oh ok I see so false and anything is always anything. That makes sense. So the F represents a substitution of the previous part but is also used to symbolize false, correct?

F=false. Presumably they would write T=true. And (false and anything)=false, that's not what they used. (false or anything)=anything.
 

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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