# Got it! Believe me, you don't want to try to figure this out. n/t

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Posted by Joel on September 26, 2002 at 02:06:46:

In Reply to: did you do a www.google.com search :) posted by Denis Borris on September 25, 2002 at 20:53:16:

: : [(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!

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