# Propositional logic question

• phospho

#### phospho

Negate ## [\neg (p\wedge \neg q)]\wedge \neg r ##

and relpace the resulting formula by an equivalent which does not involve ## \neg, \vee, \wedge ##

attempt:

## \neg ([\neg (p\wedge \neg q)]\wedge \neg r) = \neg \neg (p \wedge \neg q) \vee \neg \neg r ##

## = (p \wedge \neg q) \vee r ##
## = (p \vee r) \wedge (\neg q \vee r) = \neg ((p\vee r) \implies \neg (\neg q \vee r)) ##
## = \neg ((p \vee r) \implies \neg \neg q \wedge \neg r) = \neg ((p \vee r) \implies q \wedge \neg r) ##

## = \neg (\neg p \implies q \implies \neg (\neg q \implies r)) ##

not sure if this is correct so far, and if it is, where to go from here

Negate ## [\neg (p\wedge \neg q)]\wedge \neg r ##

and relpace the resulting formula by an equivalent which does not involve ## \neg, \vee, \wedge ##

attempt:

## \neg ([\neg (p\wedge \neg q)]\wedge \neg r) = \neg \neg (p \wedge \neg q) \vee \neg \neg r ##

## = (p \wedge \neg q) \vee r ##

You are very close at this point; try replacing ##p \wedge \neg q## with an equivalent statement not involving ## \wedge ##.

You are very close at this point; try replacing ##p \wedge \neg q## with an equivalent statement not involving ## \wedge ##.

I'm sorry I can't find one :( I'm having a lot of troubles figuring out how to find equivalent statements. I have some written down from lectures, but none which I can think of help here.

all I have is ## p \wedge \neg q = \neg (\neg p \wedge q )## but I don't know if you can "factor out" ## \neg ##

I'm sorry I can't find one :( I'm having a lot of troubles figuring out how to find equivalent statements. I have some written down from lectures, but none which I can think of help here.

Hint:
##a \wedge b \equiv \neg(a \implies \neg b)##

Hint:
##a \wedge b \equiv \neg(a \implies \neg b)##

so ## (p \wedge \neg q ) \vee q = \neg (p \implies \neg \neg q) \vee r = \neg \neg (p \implies q) \implies r) = (p \implies q) \implies r ## ?

I am very curious on how you know all of these equivalent statements? Also, are you allowed to multiply out as in algebra? e.g. does ## \neg (a \vee \neg b ) = \neg a \vee \neg \neg b ## ?

so ## (p \wedge \neg q ) \vee q = \neg (p \implies \neg \neg q) \vee r = \neg \neg (p \implies q) \implies r) = (p \implies q) \implies r ## ?

I am very curious on how you know all of these equivalent statements? Also, are you allowed to multiply out as in algebra? e.g. does ## \neg (a \vee \neg b ) = \neg a \vee \neg \neg b ## ?

You could check your answer with a truth table and see if that matches up with what you'd expect.

In general, no, the distributive law does not necessarily work in the way that you are familiar with. For instance, ##\neg(p \vee q) \neq (\neg p) \vee (\neg q)##.

You could check your answer with a truth table and see if that matches up with what you'd expect.

In general, no, the distributive law does not necessarily work in the way that you are familiar with. For instance, ##\neg(p \vee q) \neq (\neg p) \vee (\neg q)##.

hm I see,

the thing is, I see online a lot of people just stating equivalent statements, but I'm not sure how they just "know" it. I do draw truth tables, but I don't want to be doing that all the time. Is there something I'm missing?