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this is valid

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" 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

umhmm :o

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

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

uhmm :o

Help 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.

there

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

hmmm is my table okay :/ ?

I've done like three so far -_-

write

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

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

Oh okay :o... I c I c

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

neither do I

I mean, did you post them?

yeah I did ?

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...

good, sounds like you have a handle on it.