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

- anonymous

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

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- anonymous

this is valid

- 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

All four cases come out T, so the argument is valid. I think lol.

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## More answers

- anonymous

u need to verify that using a truth table?

- 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

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

"p -> ~q" and "q -> ~p" the last row has T and T so the conclusion is false?

- anonymous

actuall it mean that u shud conside the cases on when both of them are true

- anonymous

Oh I totally miss understood the question

- anonymous

So how would you do that? :o

- anonymous

help can you explain to how you do it :/

- anonymous

consider those case when p → ~q
q → ~p are true
find them

- anonymous

can you explain like in steps :/ I understand in steps.

- anonymous

hmm
consider those inputs for which
p → ~q is true
q → ~p is true

- anonymous

umhmm :o

- anonymous

lol sorry....... u shud try and understand

- anonymous

ok fine q → ~p is true for first three columns in the table

- anonymous

uhmm :o

- 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

Help D:

- anonymous

u dont need help gal
jus think

- anonymous

Blagh I am thats why I'm asking if I did right or not

- anonymous

Okay I'm guessing i didn't do that one right so ill work on another one.

- anonymous

lol draw truth tables for
p → ~q
q → ~p
and pvq see when p → ~q
q → ~p are true
pvq is true

- anonymous

(~p V q) --> ~q
(~T V F) --> ~F
(F V F) --> T
F --> T
T
So (~p V q) --> ~q is true.

- anonymous

there

- anonymous

Blagh W/e I'm right I've done everything... :l

- anonymous

q p (q → ~p) → (q ∧ ~p)
T T T
T F T
F T F
F F F

- anonymous

the first two statements are identical
\[p\to \lnot q\iff q\to \lnot p\] one is the contrapositive of the other

- anonymous

hmmm is my table okay :/ ?

- anonymous

I've done like three so far -_-

- 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

write

- 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

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

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

so none of my tables where right :/ ...

- 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

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

of course not! maybe it is sunny and i stay home anyway!

- anonymous

Oh okay :o... I c I c

- phi

I don't see the other problems..... with premises and conclusion

- anonymous

neither do I

- phi

I mean, did you post them?

- anonymous

yeah I did ?

- phi

Is this one of the problems
(~p V q) --> ~q ?
with premise ~p v q and conclusion ~q

- anonymous

I posted my answer and everything D:?

- 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

did you do this one or I did I forgot...

- 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

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

good, sounds like you have a handle on it.

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