I must show this argument using a circuit, that part I know how to do. The part i am stuck on is turning the argument into one that can be translated into a circuit
"p or q"
this changes to (p v q) ^ not q --> p
now I know i must use De Morgans Law to change the therefore but I wasn't sure which part is negated. Any help would be appreciated!
May 16th 2011, 02:43 PM
A -> B is equivalent to ~A \/ B, so (p v q) ^ not q --> p is equivalent to not((p v q) ^ not q) \/ p, or not(p \/ q) \/ q \/ p. I'll leave you to apply De Morgan one more time, if necessary. Actually, one can already see that not(p \/ q) \/ (p \/ q) is a tautology because it is the law of excluded middle.