Posted by Denis Borris on September 26, 2002 at 15:55:24:
In Reply to: Got it! Believe me, you don't want to try to figure this out. n/t posted by Joel on September 26, 2002 at 02:06:46:
: : : [(p -> q) ^ (q -> r)] -> (p -> r)
: : : This statement is a tautology. It can easily be proven to be a tautology using a truth table, but I have to prove it without using a truth table. Here's what I have (using ~ for negation, ^ for conjunction, v for disjunction, -> for implication, and <==> for equivalence):
: : : [(p -> q) ^ (q -> r)] -> (p -> r) (SAME AS ABOVE)
: : : ~[(p -> q) ^ (q -> r)] v (p -> r)
: : : [~(p -> q) v ~(q -> r)] v (p -> r)
: : : [~(~p v q) v ~(~q v r)] v (~p v r)
: : : [(p ^ ~q) v (q ^ ~r)] v (~p v r)
: : : Here I'm stuck, as far as formal proof is concerned. It's easy to see, just looking at the expression within the brackets, that if q is true, the truth value of that expression is equal to the truth value of ~r, and if q is false, the truth value of the expression is equal to the truth value of p, so somehow that expression has to boil down to (p v ~r) which is the exact contradiction of the expression on the right, thus proving the tautology. But that is not a formal proof and I can't find the steps to show that
: : : [(p ^ ~q) v (q ^ ~r)] <==> (p v ~r).
: : : Help!