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anonymous
 3 years ago
Use truth tables to test the validity of the argument.
p → ~q
q → ~p
∴ p ∨ q
anonymous
 3 years ago
Use truth tables to test the validity of the argument. p → ~q q → ~p ∴ p ∨ q

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Case p=T and q=T [(p > ~q) ^ (q > ~p)] > (p V q) [(T > ~T) ^ (T > ~T)] > (T V T) [(T > F ) ^ (T > F)] > T [ F ^ F ] > T F > T T

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0All four cases come out T, so the argument is valid. I think lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u need to verify that using a truth table?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0p q ~p ~q p > ~q q > ~p p v q T T F F F F T T F F T T T T F T T F T T T F F T T T T F Hmmm... I change my answer its invalid

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then u need to make a dumbo table with all the combinations of p and q and verify it or u can use this p>q = (~p) v q

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"p > ~q" and "q > ~p" the last row has T and T so the conclusion is false?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actuall it mean that u shud conside the cases on when both of them are true

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh I totally miss understood the question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So how would you do that? :o

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0help can you explain to how you do it :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0consider those case when p → ~q q → ~p are true find them

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you explain like in steps :/ I understand in steps.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm consider those inputs for which p → ~q is true q → ~p is true

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol sorry....... u shud try and understand

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok fine q → ~p is true for first three columns in the table

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0p q ~p ~q ∼p∨q (∼p∨q)→ ∼q ———————————————————————————————— T T F F T F T F F T F T F T T F T F F F T T T T Like this :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u dont need help gal jus think

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Blagh I am thats why I'm asking if I did right or not

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay I'm guessing i didn't do that one right so ill work on another one.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol draw truth tables for p → ~q q → ~p and pvq see when p → ~q q → ~p are true pvq is true

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(~p V q) > ~q (~T V F) > ~F (F V F) > T F > T T So (~p V q) > ~q is true.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Blagh W/e I'm right I've done everything... :l

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0q p (q → ~p) → (q ∧ ~p) T T T T F T F T F F F F

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the first two statements are identical \[p\to \lnot q\iff q\to \lnot p\] one is the contrapositive of the other

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm is my table okay :/ ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I've done like three so far _

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let me right it correctly \[\begin{array}{cccccc} P & Q & \lnot{}P & \lnot{}Q & P\to\lnot{}Q & Q\to\lnot{}P \\ \hline T & T & F & F & F & F \\ T& F & F & T & T & T \\ F & T & T & F & T & T \\ F & F & T & T & T & T \\ \hline \end{array}\]

phi
 3 years ago
Best ResponseYou've already chosen the best response.0the final step is put in the column p v q (this is the first problem) An argument is INVALID if and only if it is logically possible for the conclusion to be false even though every premise is assumed to be true.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you want to check your truth table, use this nice site here takes a second to get used to 0 and 1 instead of T and F, and a minute to learn the syntax but it is really useful http://www.kwi.dk/projects/php/truthtable/?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but try to think "logically" meaning like a human being instead of using T and F the first statement and the second statement are identical, one is the contrapositive of the other. so you do not need them both, they say the same thing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so none of my tables where right :/ ...

phi
 3 years ago
Best ResponseYou've already chosen the best response.0this looks ok, The argument is invalid because of the last row p q ~p ~q p > ~q q > ~p p v q T T F F F F T T F F T T T T F T T F T T T F F T T T T F Hmmm... I change my answer its invalid

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0P: it is raining Q: i will go to the store \(P\to \lnot Q\) if it is raining then i will not go the the store \(Q\to \lnot P\) if i go to the store, then it is not raining they are identical statements from that, can we conclude that is is raining or i go to the store?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0of course not! maybe it is sunny and i stay home anyway!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh okay :o... I c I c

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I don't see the other problems..... with premises and conclusion

phi
 3 years ago
Best ResponseYou've already chosen the best response.0Is this one of the problems (~p V q) > ~q ? with premise ~p v q and conclusion ~q

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I posted my answer and everything D:?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0because it looks invalid p q (~p V q) ~q 0 0 1 1 OK 0 1 1 0 INVALID 1 0 0 1 OK 1 1 1 0 INVALID

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you do this one or I did I forgot...

phi
 3 years ago
Best ResponseYou've already chosen the best response.0somewhere way up above you typed (~p V q) > ~q (~T V F) > ~F (F V F) > T F > T T So (~p V q) > ~q is true. I am not sure what the problem is, I'm guessing the first line. But your conclusion is not correct, because you did not test every case

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh lol thats just one of them lol. I did so many that I got lost ... but that I found out wasn't correct after all the thinking ;l

phi
 3 years ago
Best ResponseYou've already chosen the best response.0good, sounds like you have a handle on it.
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