Mathematical logic
prove the following equivalence without using truth table:
~(p v q) v (~p ^ q) = ~p

- rajat97

- jamiebookeater

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- freckles

You can make a truth table.

- rajat97

it is to be done without a truth table!

- rajat97

i'm sorry i forgot to add it

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## More answers

- rajat97

i'll add it now

- freckles

for some reason I'm thinking demorgan's law

- rajat97

yeah i did it and then tried to apply the distributive law but it was going way too long

- freckles

\[-(p \text{ or } q) \text{ or } ( -p \text{ and } q ) \\ \text {recall } -T \text{ or } -Q \equiv -(T \text{ and } Q) \\ \\ -(p \text{ or } q ) \text{ or } -(p \text{ or } - q) \\ \]
going to try to use it again
\[-([p \text{ or } q ]\text{ and } [p \text{ or } -q]) \]
I guess you can try that distributive law too
\[-(p \text{ or } (q \text{ and } -q ))\]

- freckles

I think that should make it pretty easy on you now :)

- freckles

you know since q and not q is definitely what kind of statement?

- rajat97

that's how it is solved in the textbook but i don't understand how did you use the distributive law
it should be
−([p or q] and [p or −q])
is equivalent to
-[p and (p or -q) or q and (p or -q)]
according to me

- freckles

distributive law says:
\[(P \text{ or } Q ) \text{ and } (P \text{ or } R) \equiv P \text{ or } ( Q \text{ and } R ) \\ \text{ where we have } \\ P=p \\ Q=q \\ R=-q\]
plug in

- rajat97

i may be a bit irritating but i didn't get that

- freckles

wait which part?
do you understand that is one of the distributive laws?

- freckles

like and side the not we had:
\[(p \text{ or } q ) \text{ and } ( p \text{ or } -q)\]

- freckles

this is in that exact form I mentioned above
where P=p and Q=q and R=-q

- freckles

just plug into that law above it

- freckles

\[p \text{ or } (q \text{ and } -q)\]

- freckles

and of course bring down the not sign that was in front of all of that

- freckles

like inside the not we had*

- Loser66

To me, it is quite simple
not(p or q) AND (not p AND q = (not p AND not q) AND(notp AND q)
Now, all of them are AND, we can take off the parentheses
\(\neg p \wedge \neg q\wedge \neg p \wedge q\equiv \neg p \), since \(\neg q \wedge q =0\)

- rajat97

no i didn't understand
(P or Q) and (P or R)≡P or (Q and R) where we have P=pQ=qR=−q
according to me it should be [P and (P or R)] or [Q and (P or R)]

- freckles

according to you?

- rajat97

yeah i'm sorry i mean according to the thing i read in the textbook

- freckles

oh I don't know about that rule
that what I have to be something else I try to prove
but I was using the distributive law
which says
https://en.wikipedia.org/wiki/Logical_equivalence
Instead of typing it out again

- freckles

also @Loser66 's thing looks good to me
he made it a completely an and operation thing

- freckles

she sorry

- freckles

oh that is because he changed the original statement

- freckles

she sorry again @Loser66

- rajat97

okay i'll try to get it
just a moment

- Loser66

hahaha. it's ok @freckles I worked on his/her original problem (on the very first post)

- freckles

yeah the statement would have been easier if it was and in the middle

- ganeshie8

```
~(p v q) v (~p ^ q)
(~p ^ ~q) v (~p ^ q)
~p ^ (~q v q)
~p ^ 1
~p
```

- rajat97

thanks a lot everybody for your help
we are allowed to choose only one best answer or else i would give each one of you a medal
one last question please
what if it is
(p ^ q) v (q v p)
how would the working with distributive law go?

- Loser66

=(p v (q v p)) ^ (q v (q v p))
= (p v q) ^ (q v p)
= p v q

- Loser66

=\(q\rightarrow \neg p\)

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