Results 1 to 2 of 2

Math Help - Using De Morgan's Law

  1. #1
    lrb
    lrb is offline
    Newbie
    Joined
    May 2011
    Posts
    3

    Using De Morgan's Law

    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"
    not q
    therefore p

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    769
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. De Morgan's second law Proof
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 23rd 2010, 08:07 PM
  2. Sets-Using De Morgan's Law
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 25th 2009, 08:51 AM
  3. De Morgan's laws
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: December 18th 2008, 01:16 AM
  4. de morgan's laws
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: October 24th 2008, 05:17 PM
  5. De Morgan's second law
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 7th 2008, 12:29 PM

Search Tags


/mathhelpforum @mathhelpforum