Logic: Negating if then statement

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I'm trying to negate this statement and want to make sure I'm doing it right.
(p\veeq) ---> (p \wedge q)

So I don't negate both sides do I or else that would just make them equal out again? So I just negated the left side, so \neg(p\veeq) is equivalent to \negp\wedge\negq

So that's the answer I got:
\negp\wedge\negq ---> (p \wedge q)
 
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p=>q = ~p or q

Using de morgans law : ~(p=>q) = ~(~p or q) = p and ~q

So ~(p or q implies p and q) = p or q and ~(q and p) -> p or q and ~q or ~p -> p xor q
 
Sorry I am new to all this and what you have up there is very confusing. I can't really tell what's going on and what the final answer is. Anyway you can use the actual math symbols and break the steps into separate lines? I'd really appreciate it :)
 
JonF's already done too much work for you. It's against the forum rules to simply do the problems for you. You're supposed to work them out yourself.

Use the fact that you can write p→q as (~p)∨q. The latter form is easier to see how to negate.
 
Ya you're right sorry..I was just having a hard time understanding the symbols he was using but I think I got it now..so the final answer I got was:

(p v q) ^ (~p v ~q)

Is this what you had?
 
That's correct. You can simplify it a bit if you want.
 
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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