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

[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?
Thanks in advance!

 

[fresh_42]

$p → q ⇔ \lnot \, p ∨ q$ 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.