anonymous
  • anonymous
Use truth tables to test the validity of the argument. p → ~q q → ~p ∴ p ∨ q
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
this is valid
anonymous
  • anonymous
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
anonymous
  • anonymous
All four cases come out T, so the argument is valid. I think lol.

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anonymous
  • anonymous
u need to verify that using a truth table?
anonymous
  • anonymous
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
  • anonymous
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
anonymous
  • anonymous
"p -> ~q" and "q -> ~p" the last row has T and T so the conclusion is false?
anonymous
  • anonymous
actuall it mean that u shud conside the cases on when both of them are true
anonymous
  • anonymous
Oh I totally miss understood the question
anonymous
  • anonymous
So how would you do that? :o
anonymous
  • anonymous
help can you explain to how you do it :/
anonymous
  • anonymous
consider those case when p → ~q q → ~p are true find them
anonymous
  • anonymous
can you explain like in steps :/ I understand in steps.
anonymous
  • anonymous
hmm consider those inputs for which p → ~q is true q → ~p is true
anonymous
  • anonymous
umhmm :o
anonymous
  • anonymous
lol sorry....... u shud try and understand
anonymous
  • anonymous
ok fine q → ~p is true for first three columns in the table
anonymous
  • anonymous
uhmm :o
anonymous
  • anonymous
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
anonymous
  • anonymous
Help D:
anonymous
  • anonymous
u dont need help gal jus think
anonymous
  • anonymous
Blagh I am thats why I'm asking if I did right or not
anonymous
  • anonymous
Okay I'm guessing i didn't do that one right so ill work on another one.
anonymous
  • anonymous
lol draw truth tables for p → ~q q → ~p and pvq see when p → ~q q → ~p are true pvq is true
anonymous
  • anonymous
(~p V q) --> ~q (~T V F) --> ~F (F V F) --> T F --> T T So (~p V q) --> ~q is true.
anonymous
  • anonymous
there
anonymous
  • anonymous
Blagh W/e I'm right I've done everything... :l
anonymous
  • anonymous
q p (q → ~p) → (q ∧ ~p) T T T T F T F T F F F F
anonymous
  • anonymous
the first two statements are identical \[p\to \lnot q\iff q\to \lnot p\] one is the contrapositive of the other
anonymous
  • anonymous
hmmm is my table okay :/ ?
anonymous
  • anonymous
I've done like three so far -_-
anonymous
  • anonymous
let me right it correctly \[\begin{array}{|c|c|c|c|c|c} P & Q & \lnot{}P & \lnot{}Q & P\to\lnot{}Q & Q\to\lnot{}P \\ \hline T & T & F & F & F & F \\ T& F & F & T & T & T \\ F & T & T & F & T & T \\ F & F & T & T & T & T \\ \hline \end{array}\]
anonymous
  • anonymous
write
phi
  • phi
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.
anonymous
  • anonymous
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/?
anonymous
  • anonymous
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
anonymous
  • anonymous
so none of my tables where right :/ ...
phi
  • phi
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
anonymous
  • anonymous
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?
anonymous
  • anonymous
of course not! maybe it is sunny and i stay home anyway!
anonymous
  • anonymous
Oh okay :o... I c I c
phi
  • phi
I don't see the other problems..... with premises and conclusion
anonymous
  • anonymous
neither do I
phi
  • phi
I mean, did you post them?
anonymous
  • anonymous
yeah I did ?
phi
  • phi
Is this one of the problems (~p V q) --> ~q ? with premise ~p v q and conclusion ~q
anonymous
  • anonymous
I posted my answer and everything D:?
phi
  • phi
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
anonymous
  • anonymous
did you do this one or I did I forgot...
phi
  • phi
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
anonymous
  • anonymous
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
phi
  • phi
good, sounds like you have a handle on it.

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