Maharashtra Board Class 12 Maths Part 1 Chapter 1 Mathematical Logic 1.9 Solutions

Question 1. Without using truth table, show that
(i) \( p \leftrightarrow q \equiv (p \wedge q) \vee (\sim p \wedge \sim q) \)
Answer:
LHS = \( p \leftrightarrow q \)
\( \equiv (p \rightarrow q) \wedge (q \rightarrow p) \)
\( \equiv (\sim p \vee q) \wedge (\sim q \vee p) \) .....(Conditional Law)
\( \equiv [\sim p \wedge (\sim q \vee p)] \vee [q \wedge (\sim q \vee p)] \) .....(Distributive Law)
\( \equiv [(\sim p \wedge \sim q) \vee (\sim p \wedge p)] \vee [(q \wedge \sim q) \vee (q \wedge p)] \) .....(Distributive Law)
\( \equiv [(\sim p \wedge \sim q) \vee c] \vee [c \vee (q \wedge p)] \) .....(Complement Law)
\( \equiv (\sim p \wedge \sim q) \vee (q \wedge p) \) .....(Identity Law)
\( \equiv (p \wedge q) \vee (\sim p \wedge \sim q) \) .....(Commutative Law)
\( = \text{RHS} \)
This equivalence demonstrates that a biconditional statement is true if and only if both component statements share the same truth value.
In simple words: We can prove two logical statements are the same without drawing a table by using standard logic rules step-by-step, just like simplifying an algebraic equation.

🎯 Exam Tip: Clearly write the name of the logical law used in each step in the right margin to secure full marks from the examiner.

Question. Prove the following logical equivalences:
(ii) \( p \wedge [(\sim p \vee q) \vee (\sim q)] \equiv p \)
(iii) \( \sim[(p \wedge q) \rightarrow \sim(q)] \equiv p \wedge q \)
(iv) \( \sim r \rightarrow \sim(p \wedge q) \equiv [\sim(q \rightarrow r)] \rightarrow (\sim p) \)
Answer:
(ii)
LHS \( = p \wedge [(\sim p \vee q) \vee (\sim q)] \)
\( \equiv p \wedge [\sim p \vee (q \vee \sim q)] \) ......(Associative Law)
\( \equiv p \wedge [\sim p \vee t] \) ......(Complement Law)
\( \equiv p \wedge t \) ......(Identity Law)
\( \equiv p \) ......(Identity Law)
\( = \) RHS.

(iii)
LHS \( = \sim[(p \wedge q) \rightarrow \sim(\sim q)] \)
\( \equiv (p \wedge q) \wedge \sim(\sim q) \) ......(Negation of implication)
\( \equiv (p \wedge q) \wedge q \) ......(Negation of negation)
\( \equiv p \wedge (q \wedge q) \) ......(Associative Law)
\( \equiv p \wedge q \) ......(Idempotent Law)
\( = \) RHS

(iv)
LHS \( = \sim r \rightarrow \sim(p \wedge q) \)
\( \equiv \sim q \rightarrow (\sim p \vee \sim q) \) ......(De Morgan’s Law)
\( \equiv \sim(\sim r) \vee (\sim p \vee \sim q) \) ......(Conditional Law)
\( \equiv r \vee (\sim p \vee \sim q) \) ......(Involution Law)
\( \equiv r \vee \sim q \vee \sim p \) ......(Commutative Law)
\( \equiv (\sim q \vee r) \vee (\sim p) \) ......(Commutative Law)
\( \equiv \sim(q \rightarrow r) \vee (\sim p) \) ......(Conditional Law)
\( \equiv \sim(q \rightarrow r) \rightarrow (\sim p) \) ......(Conditional Law)
\( = \) RHS.
In simple words: We can prove these logical statements are equal by applying standard rules of logic step-by-step. By simplifying the left-hand side using laws like De Morgan's or Associative laws, we show it is identical to the right-hand side.

🎯 Exam Tip: Always write down the name of the logical law used at each step in the right margin. This helps the examiner follow your logic easily and ensures you get full marks.

Question 1(v). Prove that \( (p \vee q) \rightarrow r \equiv (p \rightarrow r) \wedge (q \rightarrow r) \)
Answer:
LHS = \( (p \vee q) \rightarrow r \)
\( \equiv \sim(p \vee q) \vee r \) ...... (Conditional Law)
\( \equiv (\sim p \wedge \sim q) \vee r \) ...... (De Morgan's Law)
\( \equiv (\sim p \vee r) \wedge (\sim q \vee r) \) ...... (Distributive Law)
\( \equiv (p \rightarrow r) \wedge (q \rightarrow r) \) ...... (Conditional Law)
= RHS. This equivalence shows that a conditional statement with a disjunction in its antecedent can be split into a conjunction of two separate conditional statements.
In simple words: Saying 'if either A or B is true, then C is true' is exactly the same as saying 'if A is true then C is true, AND if B is true then C is true'.

🎯 Exam Tip: Remember that \( p \rightarrow q \) is logically equivalent to \( \sim p \vee q \). This conditional law is extremely useful for converting implications into disjunctions so you can apply De Morgan's and Distributive laws.

Question 2. Using the algebra of statement, prove that:
(i) \( [p \wedge (q \vee r)] \vee [\sim r \wedge \sim q \wedge p] \equiv p \)
(ii) \( (p \wedge q) \vee (p \wedge \sim q) \vee (\sim p \wedge \sim q) \equiv p \vee \sim q \)
Answer:
(i) LHS = \( [p \wedge (q \vee r)] \vee [\sim r \wedge \sim q \wedge p] \)
\( \equiv [p \wedge (q \vee r)] \vee [(\sim r \wedge \sim q) \wedge p] \) ...... (Associative Law)
\( \equiv [p \wedge (q \vee r)] \vee [(\sim q \wedge \sim r) \wedge p] \) ...... (Commutative Law)
\( \equiv [p \wedge (q \vee r)] \vee [\sim(q \vee r) \wedge p] \) ...... (De Morgan's Law)
\( \equiv [p \wedge (q \vee r)] \vee [p \wedge \sim(q \vee r)] \) ...... (Commutative Law)
\( \equiv p \wedge [(q \vee r) \vee \sim(q \vee r)] \) ...... (Distributive Law)
\( \equiv p \wedge t \) ...... (Complement Law)
\( \equiv p \) ...... (Identity Law)
= RHS.

(ii) LHS = \( (p \wedge q) \vee (p \wedge \sim q) \vee (\sim p \wedge \sim q) \)
\( \equiv (p \wedge q) \vee [(p \wedge \sim q) \vee (\sim p \wedge \sim q)] \) ...... (Associative Law)
\( \equiv (p \wedge q) \vee [(\sim q \wedge p) \vee (\sim q \wedge \sim p)] \) ...... (Commutative Law)
\( \equiv (p \wedge q) \vee [\sim q \wedge (p \vee \sim p)] \) ...... (Distributive Law)
\( \equiv (p \wedge q) \vee (\sim q \wedge t) \) ...... (Complement Law)
\( \equiv (p \wedge q) \vee (\sim q) \) ...... (Identity Law)
\( \equiv (p \vee \sim q) \wedge (q \vee \sim q) \) ...... (Distributive Law)
\( \equiv (p \vee \sim q) \wedge t \) ...... (Complement Law)
\( \equiv p \vee \sim q \) ...... (Identity Law)
= RHS. By systematically applying these logical laws, we can simplify complex statement formulas to their most basic equivalent forms.
In simple words: We can simplify long, complicated logical expressions step-by-step using standard rules, just like simplifying algebraic equations in math, to prove they are equal to simpler statements.

🎯 Exam Tip: Always write the name of the law used in each step in brackets on the right side. Examiners look for these specific names (like Distributive Law, Complement Law, Identity Law) to award full marks.

Question (iii). Prove that \( (p \vee q) \wedge (\sim p \vee \sim q) \equiv (p \wedge \sim q) \vee (\sim p \wedge q) \)
Answer:
LHS \( = (p \vee q) \wedge (\sim p \vee \sim q) \)
\( \equiv [p \wedge (\sim p \vee \sim q)] \vee [q \wedge (\sim p \vee \sim q)] \) ...... (Distributive Law)
\( \equiv [(p \wedge \sim p) \vee (p \wedge \sim q)] \vee [(q \wedge \sim p) \vee (q \wedge \sim q)] \) ...... (Distributive Law)
\( \equiv [c \vee (p \wedge \sim q)] \vee [(q \wedge \sim p) \vee c] \) ...... (Complement Law)
\( \equiv (p \wedge \sim q) \vee (q \wedge \sim p) \) ...... (Identity Law)
\( \equiv (p \wedge \sim q) \vee (\sim p \wedge q) \) ...... (Commutative Law)
\( = \text{RHS} \)
This step-by-step simplification successfully establishes the equivalence of both sides using standard logical laws.
In simple words: We can prove that two logical statements are the same by breaking them down step-by-step using standard rules, just like simplifying an algebra equation.

🎯 Exam Tip: Always write the name of the logical law used in brackets next to each step, as examiners look for these justifications to award full marks.