Here's the question you clicked on:
Esmy
Use truth tables to test the validity of the argument. p → ~q q → ~p ∴ p ∨ q
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
All four cases come out T, so the argument is valid. I think lol.
u need to verify that using a truth table?
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
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
"p -> ~q" and "q -> ~p" the last row has T and T so the conclusion is false?
actuall it mean that u shud conside the cases on when both of them are true
Oh I totally miss understood the question
So how would you do that? :o
help can you explain to how you do it :/
consider those case when p → ~q q → ~p are true find them
can you explain like in steps :/ I understand in steps.
hmm consider those inputs for which p → ~q is true q → ~p is true
lol sorry....... u shud try and understand
ok fine q → ~p is true for first three columns in the table
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
u dont need help gal jus think
Blagh I am thats why I'm asking if I did right or not
Okay I'm guessing i didn't do that one right so ill work on another one.
lol draw truth tables for p → ~q q → ~p and pvq see when p → ~q q → ~p are true pvq is true
(~p V q) --> ~q (~T V F) --> ~F (F V F) --> T F --> T T So (~p V q) --> ~q is true.
Blagh W/e I'm right I've done everything... :l
q p (q → ~p) → (q ∧ ~p) T T T T F T F T F F F F
the first two statements are identical \[p\to \lnot q\iff q\to \lnot p\] one is the contrapositive of the other
hmmm is my table okay :/ ?
I've done like three so far -_-
let me right it correctly \[\begin{array}{|c|c|c|c|c|c} P & Q & \lnot{}P & \lnot{}Q & P\to\lnot{}Q & Q\to\lnot{}P \\ \hline T & T & F & F & F & F \\ T& F & F & T & T & T \\ F & T & T & F & T & T \\ F & F & T & T & T & T \\ \hline \end{array}\]
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.
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/?
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
so none of my tables where right :/ ...
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
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?
of course not! maybe it is sunny and i stay home anyway!
I don't see the other problems..... with premises and conclusion
Is this one of the problems (~p V q) --> ~q ? with premise ~p v q and conclusion ~q
I posted my answer and everything D:?
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
did you do this one or I did I forgot...
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
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
good, sounds like you have a handle on it.