Dual of each of these compound propositions. I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:

a) $p∧¬q∧¬r$

b) $(p∧q∧r)∨s$

c) $(p∨F)∧(q∨T)$

My answers:

a) $¬p∨q∨r$

b) $(¬p∨¬q∨¬r)∧¬s$

c) $(¬p∧T)∨(¬q∧F)$

Did I got it right?
Thank you!
What you have done is wrong. Why you have negated all the proposition and the then find the dual?

Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).

Here is the definition of dual of a compound proposition-
"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∗." 
(Reference:  Discrete Mathematics (7th Edition) Kenneth H. Rosen)

Example:
S  =(p ∧ q)∨ (¬p ∨ q)∨ F
          S* =(p ∨ q)∧ (¬p ∧ q)∧T  (dual of S is denoted by S*)

So the correct answer to your question is:

a) p∧¬q∧¬r

b) (p∧q∧r)∨s

c) (p∨F)∧(q∨T)

answers:

a) p∨¬q∨¬r

b) (p∨q∨r)∧s

c) (p∧T)∨(q∧F)
Yes. &nbsp; All three check: okay.

You've negated all the propositions and dualled all the operators.

That is what you needed to do.