Q: Show that each condition statement is a tautology without using a truth table?
a.)
b.)
Heres the laws that I have to use:
http://www.plu.edu/~sklarjk/245s07/logequiv.pdf
Q: Show that each condition statement is a tautology without using a truth table?
a.)
b.)
Heres the laws that I have to use:
http://www.plu.edu/~sklarjk/245s07/logequiv.pdf
a) By the Distributive Law (number 3 on the list):
$\displaystyle \neg p \wedge (p\wedge q)$
Becomes:
$\displaystyle (\neg p \wedge p)\vee(\neg p \wedge q) $
$\displaystyle (\neg p \wedge p) $ is always false! They can't both be true simultaneously. Hence for the entire statement to be true, the $\displaystyle (\neg p \wedge q)$ part must be true. If this is true, then it says that $\displaystyle \neg p $ is true, and so is $\displaystyle q$. Which certainly implies that $\displaystyle q $ is true.
I see.....
I have one last question and was wondering if you can tell me if I used the right law for this.
Q: Show that ( p → q ) ^ ( p → r ) and p → ( q ^ r ) are logically equivalent?
A:
( p → q ) ^ ( p → r ) ≡ ( ¬p v q ) ^ ( ¬p v r ) Implication
≡ ¬p v ( q ^ r ) 1st Distributive Law
≡ p → (q ^ r ) Double Negation
Grillakis here is aproof based on the rules you where given ,apart the law material implication .
[~p^(pvq)] ------->q <=====>(is equivalent) ~[~p^(pvq)]vq, using material implication law.
~[~p^(pvq)]vq <=====> [pv~(pvq)]vq ,by using De Morgan
[pv~(pvq)]vq <======> qv[pv~(qvp)] ,by using commutativity twice
qv[pv~(qvp)] <======> (qvp)v~(qvp), by using associativity
(qvp)v~(qvp) <======> t , by using negation laws
I am sorry i have not figured out yet question (b)