Exercise 2.1 | Class 9 | Advanced Mathematics | Sets

1. Fill up the tables of truth values -
(A)

\(\ p \)	\(\ q \)	\(\ \sim{p} \)	\(\ \sim{q} \)	\(\ p \vee{q} \)	\(\ p \wedge{q} \)	\(\ \sim({p \vee{q})} \)	\(\ \sim{p \wedge{q}} \)	\(\ \sim{p} \vee \sim{q} \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ F \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)

(B)

\(\ p \)	\(\ q \)	\(\ \sim{p} \)	\(\ \sim{q} \)	\(\ p \wedge{\sim{q}} \)	\(\ p \to{q} \)	\(\ \sim{(p \to{q})} \)	\(\ \sim{p} \to \sim{q} \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ F \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)

(C)

\(\ p \)	\(\ q \)	\(\ p \to{q} \)	\(\ p \wedge{(p \to{q})} \)	\(\ p \wedge{(p \to{q})} \to{q} \)
\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)

2. Prove all the algebraic properties of statements with the help of truth tables.
Solution: Click Here


3. If \(\ a \), \(\ b \) and \(\ c \) are any three statements, then prove that-
(a) \(\ (a \wedge b) \to{(a \vee{b})} \)
(b) \(\ [(a \to{b}) \wedge{(b \to{c})}] \to{(a \to{c} )} \)
are two statements both of which are formula (or tentologies).
Solution:
(a)

\(\ a \)	\(\ b \)	\(\ a \wedge{b} \)	\(\ a \vee{b} \)	\(\ (a \wedge{b}) \to{(a \vee{b})} \)
\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ F \)	\(\ F \)	\(\ T \)

(b)

\(\ a \)	\(\ b \)	\(\ c \)	\(\ a \to{b} \)	\(\ b \to{c} \)	\(\ a \to{c} \)	\(\ (a \to{b}) \wedge{(b \to{c})} \)	\(\ [(a \to{b}) \wedge{(b \to{c})}] \to{(a \to{c})} \)
\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ F \)	\(\ T \)
\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)

4. Prove that
(a) \(\ \sim{(\sim{p}\wedge{\sim{q}}) \equiv p \vee q} \)
Solution:

\(\ p \)	\(\ q \)	\(\ \sim{p} \)	\(\ \sim{q} \)	\(\ p \vee{q} \)	\(\ \sim{p} \wedge{\sim{q}} \)	\(\ \sim{(\sim{p} \wedge{\sim{q}})} \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)

(b) \(\ \sim{(\sim{p}\to{\sim{q}}) \equiv \sim{p} \vee q} \)
Solution:

\(\ p \)	\(\ q \)	\(\ \sim{p} \)	\(\ \sim{q} \)	\(\ \sim{p}\to{\sim{q}} \)	\(\ \sim{(\sim{p}\to{\sim{q}})} \)	\(\ \sim{p} \wedge{q} \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ F \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)

5. Test if correct or incorrect-
(a) \(\ p \to{q} \equiv{\sim{p \to \sim{q}}} \)
Solution:

\(\ p \)	\(\ q \)	\(\ \sim{p} \)	\(\ \sim{q} \)	\(\ p \to{q} \)	\(\ \sim{p} \to{\sim{q}} \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ T \)

It is not correct.

(b) \(\ \sim{(p \to{q})} \equiv{p \to{\sim{q}}} \)
Solution:

\(\ p \)	\(\ q \)	\(\ \sim{p} \)	\(\ \sim{q} \)	\(\ p \to{q} \)	\(\ \sim{(p \to{q})} \)	\(\ p \to{\sim{q}} \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ F \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)

It is not correct.

(c) \(\ \sim{(p \to{q})} \equiv{(p \to{\sim{q}}) \vee{(q \wedge{\sim{p})}}} \)
Solution:

\(\ p \)	\(\ q \)	\(\ \sim{p} \)	\(\ \sim{q} \)	\(\ p \to{q} \)	\(\ \sim{(p \to{q})} \)	\(\ p \to{\sim{q}} \)	\(\ q \wedge{\sim{p}} \)	\(\ (p \to{\sim{q}}) \vee{(q \wedge{\sim{p})}} \)
\(\ T \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ F \)	\(\ F \)	\(\ F \)
\(\ T \)	\(\ F \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)
\(\ F \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)
\(\ F \)	\(\ F \)	\(\ T \)	\(\ T \)	\(\ T \)	\(\ F \)	\(\ T \)	\(\ F \)	\(\ T \)

It is not correct.

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