Question

Use truth tables to test the validity of the argument.

p → ~q
q → ~p
∴ p ∨ q

Answer 1

this is valid

Answer 2

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

Answer 3

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

Answer 4

u need to verify that using a truth table?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

Oh I totally miss understood the question

Answer 10

So how would you do that? :o

Answer 11

help can you explain to how you do it :/

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

umhmm :o

Answer 16

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

Answer 17

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

Answer 18

uhmm :o

Answer 19

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

Answer 20

Help D:

Answer 21

u dont need help gal
jus think

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

(~p V q) --> ~q

(~T V F) --> ~F

(F V F) --> T

F --> T

T

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

Answer 26

there

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

hmmm is my table okay :/ ?

Answer 31

I've done like three so far -_-

Answer 32

let me right it correctly
\[\begin{array}{|c|c|c|c|c|c}
P
& Q
& \lnot{}P
& \lnot{}Q
& P\to\lnot{}Q
& Q\to\lnot{}P \\
\hline
T & T & F & F & F & F \\
T& F & F & T & T & T \\
F & T & T & F & T & T \\
F & F & T & T & T & T \\
\hline
\end{array}\]

Answer 33

write

Answer 34

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.

Answer 35

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/?

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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?

Answer 40

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

Answer 41

Oh okay :o... I c I c

Answer 42

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

Answer 43

neither do I

Answer 44

I mean, did you post them?

Answer 45

yeah I did ?

Answer 46

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

Answer 47

I posted my answer and everything D:?

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

good, sounds like you have a handle on it.