Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B P → q with NOR operator and Constant F (false)

  1. Jun 16, 2016 #1
    How can I show (without using truth table) that p q is equivalent to F ↓ ((F ↓ p) ↓ q) where F is constant "false" and p and q are propositions?
    Is it possible to have a similar kind of expression with T (true) instead of F?
    Thanks in advance!
     
    Last edited: Jun 16, 2016
  2. jcsd
  3. Jun 16, 2016 #2

    fresh_42

    Staff: Mentor

    ##p → q ⇔ \lnot \, p ∨ q## and ##a \downarrow b ⇔ \lnot \, (a \, ∨\, b) = \lnot \, a \, ∧ \lnot \, b.## Thus
    $$\begin{align*}
    F \downarrow [ ( F \downarrow p) \downarrow q \, ] =&\, F \downarrow [ (\lnot \, F \, ∧ \lnot \, p) \downarrow q \,]\\
    =&\, F \downarrow [ \lnot (\lnot \, F \, ∧ \lnot \, p) ∧ \lnot q \,]\\
    =&\, F \downarrow [ (F ∨ p) ∧ \lnot q \,]\\
    =&\, \lnot F ∧ \lnot [ (F ∨ p) ∧ \lnot q \,]\\
    =&\, T ∧ [\lnot (F ∨ p) ∨ q \,]\\
    =&\, (\lnot F ∧ \lnot p ) ∨ q \,\\
    =&\, (T ∨ q) ∧ (\lnot p ∨ q) \,\\
    =&\, T ∧ (\lnot p ∨ q)\,\\
    =&\, \lnot p ∨ q\\
    =&\, (\, p → q\, )
    \end{align*}
    $$

    As ##T = \lnot \, F## you may simply substitute them.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: P → q with NOR operator and Constant F (false)
  1. If P, then Q (Replies: 6)

  2. P implies Q means ? (Replies: 1)

Loading...