MATHEMATICS
EXERCISE 1.9 (PAGE 22)
Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic
Without using truth table, show that
\[p \leftrightarrow q \equiv (p \land q) \lor (\sim p \land \sim q)\]
SOLUTION
L.H.S.
\(\quad \equiv p \leftrightarrow q\)
\(\quad \equiv (p \rightarrow q) \land (q \rightarrow p)\)
\(\quad \equiv (\sim p \lor q) \land (\sim q \lor p)\)
\(\quad \equiv [\sim p \land (\sim q \lor p)] \lor [q \land (\sim q \lor p)]\) ....[Distributive law]
\(\quad \equiv [(\sim p \land \sim q) \lor (\sim p \land p)] \lor [(q \land \sim q) \lor (q \land p)]\) ....[Distributive Law]
\(\quad \equiv [(\sim p \land \sim q) \lor \F] \lor [\F \lor (q \land p)]\) ....[Complement Law]
\(\quad \equiv (\sim p \land \sim q) \lor (q \land p)\) ....[Identity Law]
\(\quad \equiv (p \land q) \lor (\sim p \land \sim q)\) ....[Commutative Law]
\(\quad \equiv \text{R.H.S.}\)
Without using truth table, show that
\[p \land [(\sim p \lor q) \lor \sim q] \equiv p\]
SOLUTION
L.H.S.
\(\quad \equiv p \land [(\sim p \lor q) \lor \sim q]\)
\(\quad \equiv p \land [\sim p \lor (q \lor \sim q)]\) ....[Associative law]
\(\quad \equiv p \land (\sim p \lor \T)\) ....[Complement law]
\(\quad \equiv p \land \T\) ....[Identity law]
\(\quad \equiv p\) ....[Identity law]
\(\quad \equiv \text{R.H.S.}\)
Without using truth table, show that
\[\sim [(p \land q) \rightarrow \sim q] \equiv p \land q\]
SOLUTION
L.H.S.
\(\quad \equiv \sim [(p \land q) \rightarrow \sim q]\)
\(\quad \equiv (p \land q) \land \sim (\sim q)\) ....[Negation of implication, i.e., \(\sim(A \rightarrow B) \equiv A \land \sim B\)]
\(\quad \equiv (p \land q) \land q\) ....[Negation of a negation]
\(\quad \equiv p \land (q \land q)\) ....[Associative law]
\(\quad \equiv p \land q\) ....[Idempotent law, i.e., \(q \land q \equiv q\)]
\(\quad \equiv \text{R.H.S.}\)
Without using truth table, show that
\[\sim r \rightarrow \sim (p \land q) \equiv [\sim (q \rightarrow r)] \rightarrow \sim p\]
SOLUTION
L.H.S.
\(\quad \equiv \sim r \rightarrow \sim (p \land q)\)
\(\quad \equiv \sim(\sim r) \lor \sim (p \land q)\) ....[\(A \rightarrow B \equiv \sim A \lor B\)]
\(\quad \equiv r \lor \sim(p \land q)\) ....[Negation of negation]
\(\quad \equiv r \lor (\sim p \lor \sim q)\) ....[De Morgan’s law]
\(\quad \equiv \sim p \lor (\sim q \lor r)\) ....[Commutative and associative law]
\(\quad \equiv \sim p \lor (q \rightarrow r)\) ....[\(\sim A \lor B \equiv A \rightarrow B\), so \(\sim q \lor r \equiv q \rightarrow r\)]
\(\quad \equiv (q \rightarrow r) \lor \sim p\) ....[Commutative law]
\(\quad \equiv \sim[\sim (q \rightarrow r)] \lor \sim p\) ....[Negation of negation, as \(A \equiv \sim(\sim A)\)]
\(\quad \equiv [\sim (q \rightarrow r)] \rightarrow \sim p\) ....[\(\sim A \lor B \equiv A \rightarrow B\)]
\(\quad \equiv \text{R.H.S.}\)
Without using truth table, show that
\[(p \lor q) \rightarrow r \equiv (p \rightarrow r) \land (q \rightarrow r)\]
SOLUTION
L.H.S.
\(\quad \equiv (p \lor q) \rightarrow r\)
\(\quad \equiv \sim (p \lor q) \lor r\) ....[\(A \rightarrow B \equiv \sim A \lor B\)]
\(\quad \equiv (\sim p \land \sim q) \lor r\) ....[De Morgan’s law]
\(\quad \equiv (\sim p \lor r) \land (\sim q \lor r)\) ....[Distributive law]
\(\quad \equiv (p \rightarrow r) \land (q \rightarrow r)\) ....[\(\sim A \lor B \equiv A \rightarrow B\)]
\(\quad \equiv \text{R.H.S.}\)
Using the algebra of statement, prove that
\[[p \land (q \lor r)] \lor [\sim r \land \sim q \land p] \equiv p\]
SOLUTION
L.H.S.
\(\quad \equiv [p \land (q \lor r)] \lor [\sim r \land \sim q \land p]\)
\(\quad \equiv [p \land (q \lor r)] \lor [(\sim r \land \sim q) \land p]\) ....[Associative Law]
\(\quad \equiv [p \land (q \lor r)] \lor [(\sim q \land \sim r) \land p]\) ....[Commutative Law]
\(\quad \equiv [p \land (q \lor r)] \lor [\sim (q \lor r) \land p]\) ....[De Morgan’s Law]
\(\quad \equiv [p \land (q \lor r)] \lor [p \land \sim(q \lor r)]\) ....[Commutative Law]
\(\quad \equiv p \land [(q \lor r) \lor \sim(q \lor r)]\) ....[Distributive Law, \(A \land B) \lor (A \land C) \equiv A \land (B \lor C)\)]
\(\quad \equiv p \land \T\) ....[Complement Law, \(X \lor \sim X \equiv \T\)]
\(\quad \equiv p\) ....[Identity Law, \(A \land \T \equiv A\)]
\(\quad \equiv \text{R.H.S.}\)
Using the algebra of statement, prove that
\[(p \land q) \lor (p \land \sim q) \lor (\sim p \land \sim q) \equiv (p \lor \sim q)\]
SOLUTION
L.H.S.
\(\quad \equiv (p \land q) \lor (p \land \sim q) \lor (\sim p \land \sim q)\)
\(\quad \equiv [(p \land q) \lor (p \land \sim q)] \lor (\sim p \land \sim q)\) ....[Associative Law]
\(\quad \equiv [p \land (q \lor \sim q)] \lor (\sim p \land \sim q)\) ....[Distributive Law]
\(\quad \equiv (p \land \T) \lor (\sim p \land \sim q)\) ....[Complement Law]
\(\quad \equiv p \lor (\sim p \land \sim q)\) ....[Identity Law]
\(\quad \equiv (p \lor \sim p) \land (p \lor \sim q)\) ....[Distributive Law]
\(\quad \equiv \T \land (p \lor \sim q)\) ....[Complement Law]
\(\quad \equiv p \lor \sim q\) ....[Identity Law]
\(\quad \equiv \text{R.H.S.}\)
Using the algebra of statement, prove that
\[(p \lor q) \land (\sim p \lor \sim q) \equiv (p \land \sim q) \lor (\sim p \land q)\]
SOLUTION
L.H.S.
\(\quad \equiv (p \lor q) \land (\sim p \lor \sim q)\)
\(\quad \equiv [(p \lor q) \land \sim p] \lor [(p \lor q) \land \sim q]\) ....[Distributive law]
\(\quad \equiv [(p \land \sim p) \lor (q \land \sim p)] \lor [(p \land \sim q) \lor (q \land \sim q)]\) ....[Distributive law]
\(\quad \equiv [\F \lor (q \land \sim p)] \lor [(p \land \sim q) \lor \F]\) ....[Complement law]
\(\quad \equiv (q \land \sim p) \lor (p \land \sim q)\) ....[Identity law]
\(\quad \equiv (p \land \sim q) \lor (\sim p \land q)\) ....[Commutative law]
\(\quad \equiv \text{R.H.S.}\)
