SOLUTION: In need of help!!!! Use truth tables to test the validity of the argument. p → ~q q → ~p ∴ p ∨ q

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Question 588008: In need of help!!!!
Use truth tables to test the validity of the argument.
p → ~q
q → ~p
∴ p ∨ q
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Since a truth table would be so long, and you'd have to have a very wide
piece of paper to get it all in, I'd just take the four cases and work each
one out separately and see if we get T in every case:


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

Case p=T and q=F
[(p -> ~q) ^ (q -> ~p)] -> (p V q)
[(T -> ~F) ^ (F -> ~T)] -> (T V F)
[(T -> T ) ^ (F -> F) ] ->    T
[    T     ^     T    ] ->    T
           T            ->    T
                         T

Case p=F and q=T
[(p -> ~q) ^ (q -> ~p)] -> (p V q)
[(F -> ~T) ^ (T -> ~F)] -> (F V T)
[(F -> F ) ^ (T -> F )] ->    T
[    T     ^     F    ] ->    F
           F            ->    F
                         T

Case p=F and q=F
[(p -> ~q) ^ (q -> ~p)] -> (p V q)
[(F -> ~F) ^ (F -> ~F)] -> (F V F)
[(F -> T ) ^ (F -> T )] ->    F
[    T     ^     F    ] ->    F
           F            ->    F
                         T

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

Edwin