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Use truth tables to test the validity of the argument.
p → ~q
q → ~p
∴ p ∨ q
 one year ago
 one year ago
Use truth tables to test the validity of the argument. p → ~q q → ~p ∴ p ∨ q
 one year ago
 one year ago

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EsmyBest ResponseYou've already chosen the best response.1
Case 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
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
All four cases come out T, so the argument is valid. I think lol.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
u need to verify that using a truth table?
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
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
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
then 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
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
"p > ~q" and "q > ~p" the last row has T and T so the conclusion is false?
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
actuall it mean that u shud conside the cases on when both of them are true
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
Oh I totally miss understood the question
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
So how would you do that? :o
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
help can you explain to how you do it :/
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
consider those case when p → ~q q → ~p are true find them
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
can you explain like in steps :/ I understand in steps.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
hmm consider those inputs for which p → ~q is true q → ~p is true
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
lol sorry....... u shud try and understand
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
ok fine q → ~p is true for first three columns in the table
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
p 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
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
u dont need help gal jus think
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
Blagh I am thats why I'm asking if I did right or not
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
Okay I'm guessing i didn't do that one right so ill work on another one.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
lol draw truth tables for p → ~q q → ~p and pvq see when p → ~q q → ~p are true pvq is true
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
(~p V q) > ~q (~T V F) > ~F (F V F) > T F > T T So (~p V q) > ~q is true.
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
Blagh W/e I'm right I've done everything... :l
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
q p (q → ~p) → (q ∧ ~p) T T T T F T F T F F F F
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
the first two statements are identical \[p\to \lnot q\iff q\to \lnot p\] one is the contrapositive of the other
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
hmmm is my table okay :/ ?
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
I've done like three so far _
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
let me right it correctly \[\begin{array}{cccccc} P & Q & \lnot{}P & \lnot{}Q & P\to\lnot{}Q & Q\to\lnot{}P \\ \hline T & T & F & F & F & F \\ T& F & F & T & T & T \\ F & T & T & F & T & T \\ F & F & T & T & T & T \\ \hline \end{array}\]
 one year ago

phiBest ResponseYou've already chosen the best response.0
the 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.
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
if 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/?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
but 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
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
so none of my tables where right :/ ...
 one year ago

phiBest ResponseYou've already chosen the best response.0
this 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
P: 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?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
of course not! maybe it is sunny and i stay home anyway!
 one year ago

phiBest ResponseYou've already chosen the best response.0
I don't see the other problems..... with premises and conclusion
 one year ago

phiBest ResponseYou've already chosen the best response.0
I mean, did you post them?
 one year ago

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

EsmyBest ResponseYou've already chosen the best response.1
I posted my answer and everything D:?
 one year ago

phiBest ResponseYou've already chosen the best response.0
because 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
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
did you do this one or I did I forgot...
 one year ago

phiBest ResponseYou've already chosen the best response.0
somewhere 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
 one year ago

EsmyBest ResponseYou've already chosen the best response.1
Oh 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
 one year ago

phiBest ResponseYou've already chosen the best response.0
good, sounds like you have a handle on it.
 one year ago
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