1. ## Negating normal form

Hi!

I have a formula given like this(the first V is supposed to be up side down, /\ as in "and" not "or"):

$\displaystyle \sim (\sim PV \sim(\sim\sim Q \vee \sim R)) \\$

How do I approach this to "simplifie" it (negating normal form).

Any help would be greatly appriciated!

Eg.
$\displaystyle \sim (P=>Q) \\$

Which equalies/on a simpler form using De Morgan's law

$\displaystyle P.\sim Q \\$

. is "and" as in upside down V

2. Hello, jokke22!

Simplify: .$\displaystyle \sim \bigg[\sim\!P\: \wedge \sim(\sim\sim\! Q \:\vee \sim\!R)\bigg] \\$

We have: .$\displaystyle \sim\bigg[\sim\!P\:\wedge \sim(\sim\sim\!Q\:\vee \sim\!R)\bigg]$

. . . . . .$\displaystyle = \;\;\sim\bigg[P\:\wedge \sim(Q \:\vee \sim\!R)\bigg]$. . double negative

. . . . . .$\displaystyle = \;\;\sim\bigg[P\:\wedge \sim\!Q\:\wedge R\bigg]$. . . . DeMorgan's Law

. . . . . .$\displaystyle =\;\;\sim\!P \:\vee Q \:\vee \sim\!R$. . . . . DeMorgan's Law

3. This is what you posted.
$\displaystyle \begin{gathered} \neg \left[ {P \vee \neg \left( {\neg \neg Q \vee \neg R} \right)} \right] \hfill \\ \neg \left[ {P \vee \neg \left( {Q \vee \neg R} \right)} \right] \hfill \\ \neg \left[ {P \vee \left( {\neg Q \wedge R} \right)} \right] \hfill \\ \left[ {\neg P \wedge \neg \left( {\neg Q \wedge R} \right)} \right] \hfill \\ \left[ {\neg P \wedge \left( {Q \vee \neg R} \right)} \right] \hfill \\ \end{gathered}$

Did you post it correctly?

4. Appriciate it mate! Thank you!

Since I have your attention, im struggling a bit with this one as well:

$\displaystyle \sim\bigg[(P->Q)->\sim(R->S)\bigg]$

5. Originally Posted by Plato
This is what you posted.
$\displaystyle \begin{gathered} \neg \left[ {P \vee \neg \left( {\neg \neg Q \vee \neg R} \right)} \right] \hfill \\ \neg \left[ {P \vee \neg \left( {Q \vee \neg R} \right)} \right] \hfill \\ \neg \left[ {P \vee \left( {\neg Q \wedge R} \right)} \right] \hfill \\ \left[ {\neg P \wedge \neg \left( {\neg Q \wedge R} \right)} \right] \hfill \\ \left[ {\neg P \wedge \left( {Q \vee \neg R} \right)} \right] \hfill \\ \end{gathered}$

Did you post it correctly?
Yes that is correct, EXCEPT for the first $\displaystyle \vee$ that should be $\displaystyle \wedge$, sorry for my bad typing here, not familiar with all the commands yet!

6. Originally Posted by jokke22
Appriciate it mate! Thank you!

Since I have your attention, im struggling a bit with this one as well:

$\displaystyle \sim\bigg[(P->Q)->\sim(R->S)\bigg]$

~[(p--->q)----->~(r---->s)] =

=~[~(~pvq) v ~(~r v s)]=..................by material implication:P---->Q=~P V Q

= (~pvq) ^ (~r v s)=.........................by de morgan...

=(p---->q) ^ (r----->s).....................by material implication again

7. Hello again, jokke22!

$\displaystyle \sim\bigg[(P \to Q)\;\to\; \sim\!(R \to S)\bigg]$

$\displaystyle \sim\bigg[(\sim\! P \vee Q) \;\to\;\sim(\sim\! R \vee S)\bigg]$ . . def. of Implication

$\displaystyle \sim\bigg[(\sim\!P \vee Q) \;\to\; (R\: \wedge \sim\!S)\bigg]$. . . .DeMorgan

$\displaystyle \sim\bigg[\sim(\sim\!P \vee Q) \;\vee\; (R \:\wedge \sim\!S)\bigg]$ . . def. of Implication

. . . $\displaystyle (\sim\!P \vee Q) \;\wedge\; \sim(R \:\wedge \sim\!S)$ . . .DeMorgan

. . . $\displaystyle (\sim\!P \vee Q) \;\wedge\; (\sim\!R \vee S)$ . . . . DeMorgan

8. Originally Posted by Soroban
Hello again, jokke22!

$\displaystyle \sim\bigg[(\sim\! P \vee Q) \;\to\;\sim(\sim\! R \vee S)\bigg]$ . . def. of Implication

$\displaystyle \sim\bigg[(\sim\!P \vee Q) \;\to\; (R\: \wedge \sim\!S)\bigg]$. . . .DeMorgan

$\displaystyle \sim\bigg[\sim(\sim\!P \vee Q) \;\vee\; (R \:\wedge \sim\!S)\bigg]$ . . def. of Implication

. . . $\displaystyle (\sim\!P \vee Q) \;\wedge\; \sim(R \:\wedge \sim\!S)$ . . .DeMorgan

. . . $\displaystyle (\sim\!P \vee Q) \;\wedge\; (\sim\!R \vee S)$ . . . . DeMorgan

Thank you yet again! As to benes!
Now I see how these can be solved by adding these laws one by one to simplify them!
As to the laws and such do you know of any great resources which I can look at as well?

9. As for this:

$\displaystyle \sim\bigg[$$\displaystyle \sim (\ P \to Q) \vee\sim R \bigg] \displaystyle \sim\bigg[$$\displaystyle \sim (\sim \ P \vee Q) \vee\sim R \bigg]$

$\displaystyle \sim\bigg[$$\displaystyle P \vee (Q \vee\sim R) \bigg] \displaystyle \bigg[$$\displaystyle \sim P \wedge (Q \vee\sim R) \bigg]$

$\displaystyle \sim P \wedge (\sim Q \wedge R)$

Gave it a try but pretty sure this one is wrong though...