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

1. Jun 16, 2016

### SamitC

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?

Last edited: Jun 16, 2016
2. Jun 16, 2016

### 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.