rajat97
  • rajat97
Mathematical logic prove the following equivalence without using truth table: ~(p v q) v (~p ^ q) = ~p
Mathematics
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
You can make a truth table.
rajat97
  • rajat97
it is to be done without a truth table!
rajat97
  • rajat97
i'm sorry i forgot to add it

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rajat97
  • rajat97
i'll add it now
freckles
  • freckles
for some reason I'm thinking demorgan's law
rajat97
  • rajat97
yeah i did it and then tried to apply the distributive law but it was going way too long
freckles
  • 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
  • freckles
I think that should make it pretty easy on you now :)
freckles
  • freckles
you know since q and not q is definitely what kind of statement?
rajat97
  • 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
  • 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
  • rajat97
i may be a bit irritating but i didn't get that
freckles
  • freckles
wait which part? do you understand that is one of the distributive laws?
freckles
  • freckles
like and side the not we had: \[(p \text{ or } q ) \text{ and } ( p \text{ or } -q)\]
freckles
  • freckles
this is in that exact form I mentioned above where P=p and Q=q and R=-q
freckles
  • freckles
just plug into that law above it
freckles
  • freckles
\[p \text{ or } (q \text{ and } -q)\]
freckles
  • freckles
and of course bring down the not sign that was in front of all of that
freckles
  • freckles
like inside the not we had*
Loser66
  • 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
  • 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
  • freckles
according to you?
rajat97
  • rajat97
yeah i'm sorry i mean according to the thing i read in the textbook
freckles
  • 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
  • freckles
also @Loser66 's thing looks good to me he made it a completely an and operation thing
freckles
  • freckles
she sorry
freckles
  • freckles
oh that is because he changed the original statement
freckles
  • freckles
she sorry again @Loser66
rajat97
  • rajat97
okay i'll try to get it just a moment
Loser66
  • Loser66
hahaha. it's ok @freckles I worked on his/her original problem (on the very first post)
freckles
  • freckles
yeah the statement would have been easier if it was and in the middle
ganeshie8
  • ganeshie8
``` ~(p v q) v (~p ^ q) (~p ^ ~q) v (~p ^ q) ~p ^ (~q v q) ~p ^ 1 ~p ```
rajat97
  • 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
  • Loser66
=(p v (q v p)) ^ (q v (q v p)) = (p v q) ^ (q v p) = p v q
Loser66
  • Loser66
=\(q\rightarrow \neg p\)

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