Q: Show that each condition statement is a tautology without using a truth table?
Heres the laws that I have to use:
is always false! They can't both be true simultaneously. Hence for the entire statement to be true, the part must be true. If this is true, then it says that is true, and so is . Which certainly implies that is true.
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?
( p → q ) ^ ( p → r ) ≡ ( ¬p v q ) ^ ( ¬p v r ) Implication
≡ ¬p v ( q ^ r ) 1st Distributive Law
≡ p → (q ^ r ) Double Negation
[~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)