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Find the dual of each of these compound propositions.

1. $p \wedge \sim q \wedge \sim r$
2. $(p \wedge q \wedge r) \vee s$
3. $(p \vee F) \wedge (q \vee T)$
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Dual is same as de morgans low . Except we can not complement  of literls in dual .

answered by Boss (33.9k points)
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Literal$:-$  every occurrence of a variable in its true form $(or)$ complemented form.

Example$: A.\overline{B}+\overline{A}.B$

Number of literals$=4$

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Dual = "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∗."

we don't negate any of the literal.

Answer:

a)  p$\vee$∼q$\vee$∼r

b)  (p$\vee$q$\vee$r)$\wedge$s

c)  (p$\wedge$T)$\vee$(q$\wedge$F)

answered by Active (1.3k points)

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