When does a dual of a compound proposition equal itself? So I am studying computer science and right now I am stuck on a problem.
When does s∗ = s, where s is a compound proposition?
So far the only thing I can come up with is:
s* = s when the compound proposition is composed only of the same           propositions. (ex. p ∧ p = p ∨ p)
The book defines duality as:
The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗.
(Discrete Mathematics and its Applications, Rosen, 7e)
Any help would be great, this is a tricky one.
 A: I hope this is going to help somebody; 
I came up with the same solution which is known as idempotent law, plus the others:
 p ∧ T = p ∨ F (identity law)
 p∧(p∨q)=p∨(p∧q) (absorption law)  (try proving logical equivalences) 
A: Even I use Discrete Mathematics by Rosen ,7r.
Duality & its really pretty awesome book for Discrete mathematics.
The dual of a compound proposition that contains only the logical operators of AND (^) ,OR (v) and negation (~) is the compound proposition obtained by replacing each v by ^, each ^ by v, each T by F,and each F by T. The dual of of a compound proposition s is denoted by s*.
so we will have dual s = s* & you can verify it by using truth tables.
For. e.g:
p ^ p  is equivalent to its dual p v p and as you construct a truth table for those duals, you'll see that both are logically equal. So, s = s* = s.
Similar is for (s*)* = s
 i.e.,
as we know that a dual is formed by replacing And with an OR and vice-versa is also possible and same is the case when dealing with T and F .
To be precise, a dual for p v p is p^p   i.e., s*  is dual of s and again if you apply the duality law for s*  again, you'll see that you get back s again i.e.,s = s* = (s*)* = s.
