# Duality discrete math problem. 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∗. Find the dual of these compound propositions.

Duality discrete math problem
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∗. Find the dual of these compound propositions.
a) $p\vee \mathrm{¬}q$
I got $\mathrm{¬}p\wedge q$
b) $p\wedge \left(q\vee \left(r\wedge \mathrm{T}\right)\right)$
My answer was $\mathrm{¬}p\vee \left(\mathrm{¬}q\wedge r\right)$
c) $\left(p\wedge \mathrm{¬}q\right)\vee \left(q\wedge \mathrm{F}\right)$
My answer was $\left(\mathrm{¬}p\vee q\right)\wedge \mathrm{¬}q$
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Tristan Pittman
Step 1
You did more as you should. Forming the dual just wants you to replace p by $\mathrm{¬}p$ for each literal p, ∨ by ∧ and vice versa and T by F. You did more than that, in dualising (2), one obtains
$\mathrm{¬}p\vee \left(\mathrm{¬}q\wedge \left(\mathrm{¬}r\vee F\right)\right)$
Step 2
(you missed a $\mathrm{¬}$ in front of r). We have of course $\mathrm{¬}r\vee F\equiv \mathrm{¬}r$, but this is not part of dualising. Same for (3), the dual proposition is
$\left(\mathrm{¬}p\vee q\right)\wedge \left(\mathrm{¬}q\vee T\right)$
###### Not exactly what you’re looking for?
Karsyn Beltran
Step 1
What you have done is wrong. Why you have negated all the proposition and the then find the dual?
Like dual of
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∗.
Step 2
Example: $S=\left(p\wedge q\right)\vee \left(¬p\vee q\right)\vee F$
dual of $S=\left(p\vee q\right)\wedge \left(¬p\wedge q\right)\wedge T$
a) $p\vee ¬q$
dual: $p\wedge ¬q$
b) $p\wedge \left(q\vee \left(r\wedge T\right)\right)$
dual: $p\vee \left(q\wedge \left(r\vee F\right)\right)$
c) $\left(p\wedge ¬q\right)\vee \left(q\wedge F\right)$
dual: $\left(p\vee ¬q\right)\wedge \left(q\vee T\right)$