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

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  • Thread starter SamitC
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  • #1
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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!
 
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Answers and Replies

  • #2
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##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.
 

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