Logical Equivalence of x <=> y and (x-->y) ^ ((~x)-->(~y))

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Prove x <=> y is logically equivalent to (x-->y) ^ ((~x)-->(~y)).
 
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aorick21 said:
Prove x <=> y is logically equivalent to (x-->y) ^ ((~x)-->(~y)).

Please show some sort of work or at least tell us where you are stuck. We help with your homework, not do your homework.
 
Actually, x \Leftrightarrow y is shorthand for a longer expression. Which one? Now rewrite one of the subexpressions and you're done.

If you want more specific help, please refer to l46kok's post.

Also, how specific do you need the proof to be? Can you use "intuitive" rules or do you really have to produce a proof tree?
 
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I'd just use a truth table.

If you don't know what that is, then I don't think you belong in math.
 
TimNguyen said:
I'd just use a truth table.

If you don't know what that is, then I don't think you belong in math.

Rubbish. If you do know what one is then perhaps you belong in computer science or electrical engineering?
 
Aoik: what is the contrapositive of ~x=>~y

p.s.: Matt, you crack me up.
 

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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