Suppose, P$_{1}$ & P$_{2}$ are two equivalent compound proposition.
P$_{1}$ $\equiv$ P$_{2}$
then complement of these compound proposition is also equivalent.
$\neg$P$_{1}$ $\equiv\neg$P$_{2}$
for e.g. : Suppose we have a function $f_{(a,b)}$.we make Truth Table of $f_{(a,b)}$ ,
| a |
b |
P$_{1}$ |
P$_{2}$ |
$\neg$P$_{1}$ |
$\neg$P$_{2}$ |
| F |
F |
F |
F |
T |
T |
| F |
T |
F |
F |
T |
T |
| T |
F |
T |
T |
F |
F |
| T |
T |
F |
F |
T |
T |
from the above truth table we can prove that if P$_{1} \equiv$ P$_{2}$ then $\neg$P$_{1} \equiv \neg$P$_{2}$.
Complement of a function ($f_{c}$): complement of all literal, $\wedge$ changes to $\vee$, $\vee$ changes to $\wedge$, T to F, F to T.
Dual of a function ($f^{*}$): $\wedge$ changes to $\vee$, $\vee$ changes to $\wedge$, T to F, F to T.
Now, from the above definitions we can say that if $f$ is a function then complement of a function $f_{c}$ is equivalent to its dual $f^{*}$ by complementing of every literal of $f_{c}$ that means $f_{c} \equiv f^{*}$.
Hence, from $\neg$P$_{1}$ $\equiv\neg$P$_{2}$ we can conclude P$^{*}_{1} \equiv$ P$^{*}_{2}$.
