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Detailed Chapter 1 Mathematical Logic 1.6 MSBSHSE Solutions for Class 12 Maths Commerce
For Class 12 students, solving MSBSHSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 12 Maths Commerce solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 1 Mathematical Logic 1.6 solutions will improve your exam performance.
Class 12 Maths Commerce Chapter 1 Mathematical Logic 1.6 MSBSHSE Solutions PDF
Question 1. Prepare the truth tables for the following statement patterns:
(i) \( p \rightarrow (\sim p \vee q) \)
Answer:
Here are two statements and three connectives. The truth table for the statement pattern \( p \rightarrow (\sim p \vee q) \) is constructed below:
| \( p \) | \( q \) | \( \sim p \) | \( \sim p \vee q \) | \( p \rightarrow (\sim p \vee q) \) |
|---|---|---|---|---|
| T | T | F | T | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | T | T |
In simple words: A truth table helps us see the final truth value of a statement for every possible combination of true and false inputs. We solve it step-by-step, just like a math equation.
🎯 Exam Tip: Draw the truth table columns clearly and double-check the conditional (\( \rightarrow \)) column, remembering that it is only false when True implies False.
Question 1. Construct the truth table for the statement pattern: \( p \rightarrow (\sim p \vee q) \)
Answer:
There are 2 statements and 3 connectives.
\( \implies \) there are \( 2 \times 2 = 4 \) rows and \( 2 + 3 = 5 \) columns in the truth table.
| \( p \) | \( q \) | \( \sim p \) | \( \sim p \vee q \) | \( p \rightarrow (\sim p \vee q) \) |
|---|---|---|---|---|
| T | T | F | T | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | T | T |
In simple words: This truth table shows all possible combinations of truth values for statements p and q, helping us find the final truth value of the entire expression step-by-step.
🎯 Exam Tip: Clearly list the number of rows and columns before drawing the table to show your step-by-step understanding and avoid missing any columns.
Question 2. Construct the truth table for the statement pattern: \( (\sim p \vee q) \wedge (\sim p \vee \sim q) \)
Answer:
There are 2 statements and 5 connectives.
\( \implies \) there are \( 2 \times 2 = 4 \) rows and \( 2 + 5 = 7 \) columns in the truth table.
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( \sim p \vee q \) | \( \sim p \vee \sim q \) | \( (\sim p \vee q) \wedge (\sim p \vee \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | T | F | F |
| T | F | F | T | F | T | F |
| F | T | T | F | T | T | T |
| F | F | T | T | T | T | T |
In simple words: We break down the complex statement into smaller parts like negations and disjunctions, then combine them using conjunction to find the final truth values.
🎯 Exam Tip: Double-check the truth values for conjunction (\( \wedge \)) and disjunction (\( \vee \)) operations to ensure no simple logical errors are made in intermediate columns.
Question 3. Construct the truth table for the statement pattern: \( (p \wedge r) \rightarrow (p \vee \sim q) \)
Answer:
Here are three statements and 4 connectives.
\( \implies \) there are \( 2 \times 2 \times 2 = 8 \) rows and \( 3 + 4 = 7 \) columns in the truth table.
| \( p \) | \( q \) | \( r \) | \( \sim q \) | \( p \wedge r \) | \( p \vee \sim q \) | \( (p \wedge r) \rightarrow (p \vee \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | T | F | T | T | T |
| T | T | F | F | F | T | T |
| T | F | T | T | T | T | T |
| T | F | F | T | F | T | T |
| F | T | T | F | F | F | T |
| F | T | F | F | F | F | T |
| F | F | T | T | F | T | T |
| F | F | F | T | F | T | T |
In simple words: When we have three statements, we need 8 rows to cover all possible combinations of True and False, and we solve each part step-by-step to find the final implication.
🎯 Exam Tip: For 3-variable truth tables, write the first column as 4 Ts and 4 Fs, the second as alternating pairs of Ts and Fs, and the third as alternating single Ts and Fs to ensure you don't miss any combination.
Question 1. (iv) \( (p \wedge q) \vee \sim r \)
Answer:
The truth table for the given statement pattern is constructed below:
| \( p \) | \( q \) | \( r \) | \( \sim r \) | \( p \wedge q \) | \( (p \wedge q) \vee \sim r \) |
|---|---|---|---|---|---|
| T | T | T | F | T | T |
| T | T | F | T | T | T |
| T | F | T | F | F | F |
| T | F | F | T | F | T |
| F | T | T | F | F | F |
| F | T | F | T | F | T |
| F | F | T | F | F | F |
| F | F | F | T | F | T |
This truth table systematically evaluates the truth values of the compound statement for all possible combinations of its components.
In simple words: This table helps us find the final truth value of the statement for every possible combination of true and false values of p, q, and r. We solve it step-by-step, just like a math equation.
🎯 Exam Tip: Double-check the negation column (\( \sim r \)) first, as a simple mistake there will affect the final column's values.
Question 2. Examine, whether each of the following statement patterns is a tautology or a contradiction or a contingency:
(i) \( q \vee [\sim(p \wedge q)] \)
(ii) \( (\sim q \wedge p) \wedge (p \wedge \sim p) \)
Answer:
(i) \( q \vee [\sim(p \wedge q)] \)
The truth table is as follows:
| \( p \) | \( q \) | \( p \wedge q \) | \( \sim(p \wedge q) \) | \( q \vee [\sim(p \wedge q)] \) |
|---|---|---|---|---|
| T | T | T | F | T |
| T | F | F | T | T |
| F | T | F | T | T |
| F | F | F | T | T |
All the entries in the last column of the above truth table are T. Therefore, \( q \vee [\sim(p \wedge q)] \) is a tautology. This confirms that the statement is universally true under all logical interpretations.
(ii) \( (\sim q \wedge p) \wedge (p \wedge \sim p) \)
The truth table is as follows:
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( \sim q \wedge p \) | \( p \wedge \sim p \) | \( (\sim q \wedge p) \wedge (p \wedge \sim p) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | F | F | F |
| T | F | F | T | T | F | F |
| F | T | T | F | F | F | F |
| F | F | T | T | F | F | F |
All the entries in the last column of the above truth table are F. Therefore, \( (\sim q \wedge p) \wedge (p \wedge \sim p) \) is a contradiction. This shows that the statement can never be true under any circumstances.
In simple words: A tautology is a statement that is always true (all Ts in the final column), while a contradiction is always false (all Fs in the final column). By building these tables, we proved the first one is a tautology and the second one is a contradiction.
🎯 Exam Tip: Always write a concluding sentence stating whether the pattern is a tautology, contradiction, or contingency based on the final column to secure full marks.
Question 2. State whether each of the following statement patterns is a tautology, contradiction, or contingency:
(iii) \( (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \)
(iv) \( \sim p \rightarrow (p \rightarrow \sim q) \)
Answer:
(iii) Truth table for \( (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \):
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \wedge \sim q \) | \( \sim p \wedge \sim q \) | \( (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | F | F | T |
| T | F | F | T | T | F | F |
| F | T | T | F | F | F | T |
| F | F | T | T | F | T | T |
The entries in the last column are neither all T nor all F. Therefore, the statement pattern \( (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \) is a contingency.
(iv) Truth table for \( \sim p \rightarrow (p \rightarrow \sim q) \):
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \rightarrow \sim q \) | \( \sim p \rightarrow (p \rightarrow \sim q) \) |
|---|---|---|---|---|---|
| T | T | F | F | F | T |
| T | F | F | T | T | T |
| F | T | T | F | T | T |
| F | F | T | T | T | T |
All the entries in the last column of the truth table are T. Therefore, the statement pattern \( \sim p \rightarrow (p \rightarrow \sim q) \) is a tautology. This classification helps us understand the logical behavior of complex statements under all possible truth value assignments.
In simple words: A statement pattern is a contingency if its truth table contains a mix of true and false values, whereas it is a tautology if every single row in the final column results in true.
🎯 Exam Tip: When constructing truth tables, always write down the basic truth values for p and q systematically to avoid missing any combinations.
Question 3. Prove that each of the following statement pattern is a tautology:
(i) \( (p \wedge q) \rightarrow q \)
Answer:
(i) Truth table for \( (p \wedge q) \rightarrow q \):
| \( p \) | \( q \) | \( p \wedge q \) | \( (p \wedge q) \rightarrow q \) |
|---|---|---|---|
| T | T | T | T |
| T | F | F | T |
| F | T | F | T |
| F | F | F | T |
All the entries in the last column of the truth table are T. Therefore, the statement pattern \( (p \wedge q) \rightarrow q \) is a tautology. A tautology is a fundamental concept in mathematical logic that represents a statement which is universally true.
In simple words: Since the final column of the truth table contains only True (T) values, the statement is always true under all conditions, making it a tautology.
🎯 Exam Tip: When proving a tautology, ensure that every single row in your final column ends with 'T' and clearly state this observation in your final concluding sentence.
Question (ii) \( (p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p) \)
Answer:
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \rightarrow q \) | \( \sim q \rightarrow \sim p \) | \( (p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | T | T | T |
| T | F | F | T | F | F | T |
| F | T | T | F | T | T | T |
| F | F | T | T | T | T | T |
All the entries in the last column of the above truth table are T. This confirms that the statement is logically valid under all possible truth assignments.
\( \therefore (p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p) \) is a tautology.
In simple words: Since the final column of our truth table contains only 'T' (True) values, the statement is always true under all conditions, making it a tautology.
🎯 Exam Tip: Double-check the truth values for conditional (\( \rightarrow \)) and biconditional (\( \leftrightarrow \)) operators carefully, as these are common areas where simple mistakes occur.
Question (iii) \( (\sim p \wedge \sim q) \rightarrow (p \rightarrow q) \)
Answer:
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( \sim p \wedge \sim q \) | \( p \rightarrow q \) | \( (\sim p \wedge \sim q) \rightarrow (p \rightarrow q) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | F | T | T |
| T | F | F | T | F | F | T |
| F | T | T | F | F | T | T |
| F | F | T | T | T | T | T |
All the entries in the last column of the above truth table are T. This demonstrates that the conditional relationship holds true across all scenarios.
\( \therefore (\sim p \wedge \sim q) \rightarrow (p \rightarrow q) \) is a tautology.
In simple words: A statement is a tautology when every single combination of truth values results in a 'True' outcome in the final column.
🎯 Exam Tip: When constructing truth tables, write down the columns step-by-step from simpler sub-expressions to the final complex statement to avoid confusion.
Question (iv) \( (\sim p \vee \sim q) \leftrightarrow \sim(p \wedge q) \)
Answer:
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( \sim p \vee \sim q \) | \( p \wedge q \) | \( \sim(p \wedge q) \) | \( (\sim p \vee \sim q) \leftrightarrow \sim(p \wedge q) \) |
|---|---|---|---|---|---|---|---|
| T | T | F | F | F | T | F | T |
| T | F | F | T | T | F | T | T |
| F | T | T | F | T | F | T | T |
| F | F | T | T | T | F | T | T |
All the entries in the last column of the above truth table are T. This equivalence shows that both sides of the biconditional represent the exact same logical meaning.
\( \therefore (\sim p \vee \sim q) \leftrightarrow \sim(p \wedge q) \) is a tautology.
In simple words: This truth table proves De Morgan's Law, showing that saying 'not A or not B' is logically identical to saying 'not (A and B)'.
🎯 Exam Tip: For biconditional (\( \leftrightarrow \)) statements, the final column will be 'T' only if both compared columns have identical truth values in each row.
Question 4. Prove that each of the following statement pattern is a contradiction:
(i) \( (p \vee q) \wedge (\sim p \wedge \sim q) \)
(ii) \( (p \wedge q) \wedge \sim p \)
(iii) \( (p \wedge q) \wedge (\sim p \vee \sim q) \)
Answer:
(i) Truth table for \( (p \vee q) \wedge (\sim p \wedge \sim q) \):
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \vee q \) | \( \sim p \wedge \sim q \) | \( (p \vee q) \wedge (\sim p \wedge \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | T | F | F |
| T | F | F | T | T | F | F |
| F | T | T | F | T | F | F |
| F | F | T | T | F | T | F |
All the entries in the last column of the above truth table are F. Since all truth values in the final column are false, the statement pattern is false under all possible truth value assignments of its components.
\( \therefore (p \vee q) \wedge (\sim p \wedge \sim q) \) is a contradiction.
(ii) Truth table for \( (p \wedge q) \wedge \sim p \):
| \( p \) | \( q \) | \( \sim p \) | \( p \wedge q \) | \( (p \wedge q) \wedge \sim p \) |
|---|---|---|---|---|
| T | T | F | T | F |
| T | F | F | F | F |
| F | T | T | F | F |
| F | F | T | F | F |
All the entries in the last column of the above truth table are F.
\( \dots (p \wedge q) \wedge \sim p \) is a contradiction.
(iii) Truth table for \( (p \wedge q) \wedge (\sim p \vee \sim q) \):
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \wedge q \) | \( \sim p \vee \sim q \) | \( (p \wedge q) \wedge (\sim p \vee \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | T | F | F |
| T | F | F | T | F | T | F |
| F | T | T | F | F | T | F |
| F | F | T | T | F | T | F |
All the entries in the last column of the above truth table are F.
\( \therefore (p \wedge q) \wedge (\sim p \vee \sim q) \) is a contradiction.
In simple words: A contradiction is a statement that is always false, no matter what the starting values of p and q are. By drawing these truth tables, we can see that the final column contains only 'F' (False) in every single row, proving they are indeed contradictions.
🎯 Exam Tip: Double-check your truth table columns step-by-step, especially negation and conjunction operations, to ensure no simple calculation errors lead to an incorrect final column.
Question 4. (iv) \( (p \rightarrow q) \wedge (p \wedge \sim q) \)
Answer:
| \( p \) | \( q \) | \( \sim q \) | \( p \rightarrow q \) | \( p \wedge \sim q \) | \( (p \rightarrow q) \wedge (p \wedge \sim q) \) |
|---|---|---|---|---|---|
| T | T | F | T | F | F |
| T | F | T | F | T | F |
| F | T | F | T | F | F |
| F | F | T | T | F | F |
All the entries in the last column of the above truth table are F.
\( \therefore (p \rightarrow q) \wedge (p \wedge \sim q) \) is a contradiction.
In simple words: Since the final column of the truth table contains only 'F' (False) values, the statement is always false, making it a contradiction.
🎯 Exam Tip: When proving a contradiction, double-check that every single entry in the final column of your truth table is 'F'.
Question 5. Show that each of the following statement pattern is a contingency:
(i) \( (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \)
(ii) \( (p \rightarrow q) \leftrightarrow (\sim p \wedge q) \)
Answer:
(i) \( (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \)
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \wedge \sim q \) | \( \sim p \wedge \sim q \) | \( (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \) |
|---|---|---|---|---|---|---|
| T | T | F | F | F | F | T |
| T | F | F | T | T | F | F |
| F | T | T | F | F | F | T |
| F | F | T | T | F | T | T |
The entries in the last column of the above truth table are neither all T nor all F.
\( \therefore (p \wedge \sim q) \rightarrow (\sim p \wedge \sim q) \) is a contingency.
(ii) \( (p \rightarrow q) \leftrightarrow (\sim p \wedge q) \)
| \( p \) | \( q \) | \( \sim p \) | \( p \rightarrow q \) | \( \sim p \wedge q \) | \( (p \rightarrow q) \leftrightarrow (\sim p \wedge q) \) |
|---|---|---|---|---|---|
| T | T | F | T | F | F |
| T | F | F | F | F | T |
| F | T | T | T | T | T |
| F | F | T | T | F | F |
The entries in the last column of the above truth table are neither all T nor all F.
\( \therefore (p \rightarrow q) \leftrightarrow (\sim p \wedge q) \) is a contingency.
In simple words: A contingency is a statement that can be either true or false depending on the situation. Since the final columns of both tables contain a mix of T and F, both statement patterns are contingencies.
🎯 Exam Tip: Remember that a contingency must have at least one 'T' and at least one 'F' in its final column. Clearly state this observation to secure full marks.
Question 1. Classify the following statement patterns as a tautology, contradiction or contingency:
(ii) \( (p \rightarrow q) \leftrightarrow (\sim p \wedge q) \)
(iii) \( p \wedge [(p \rightarrow \sim q) \rightarrow q] \)
(iv) \( (p \rightarrow q) \wedge (p \rightarrow r) \)
Answer:
(ii) \( (p \rightarrow q) \leftrightarrow (\sim p \wedge q) \)
| \( p \) | \( q \) | \( \sim p \) | \( p \rightarrow q \) | \( \sim p \wedge q \) | \( (p \rightarrow q) \leftrightarrow (\sim p \wedge q) \) |
|---|---|---|---|---|---|
| T | T | F | T | F | F |
| T | F | F | F | F | T |
| F | T | T | T | T | T |
| F | F | T | T | F | F |
The entries in the last column of the above truth table are neither all T nor all F.
\(\therefore (p \rightarrow q) \leftrightarrow (\sim p \wedge q)\) is a contingency.
(iii) \( p \wedge [(p \rightarrow \sim q) \rightarrow q] \)
| \( p \) | \( q \) | \( \sim q \) | \( p \rightarrow \sim q \) | \( (p \rightarrow \sim q) \rightarrow q \) | \( p \wedge [(p \rightarrow \sim q) \rightarrow q] \) |
|---|---|---|---|---|---|
| T | T | F | F | T | T |
| T | F | T | T | F | F |
| F | T | F | T | T | F |
| F | F | T | T | F | F |
The entries in the last column of the above truth table are neither all T nor all F.
\(\therefore p \wedge [(p \rightarrow \sim q) \rightarrow q]\) is a contingency.
(iv) \( (p \rightarrow q) \wedge (p \rightarrow r) \)
| \( p \) | \( q \) | \( r \) | \( p \rightarrow q \) | \( p \rightarrow r \) | \( (p \rightarrow q) \wedge (p \rightarrow r) \) |
|---|---|---|---|---|---|
| T | T | T | T | T | T |
| T | T | F | T | F | F |
| T | F | T | F | T | F |
| T | F | F | F | F | F |
| F | T | T | T | T | T |
| F | T | F | T | T | T |
| F | F | T | T | T | T |
| F | F | F | T | T | T |
The entries in the last column of the above truth table are neither all T nor all F.
\(\therefore (p \rightarrow q) \wedge (p \rightarrow r)\) is a contingency.
In simple words: A statement pattern is called a contingency if its final truth values are a mix of true (T) and false (F). Since all three tables have both T and F in their final columns, they are all classified as contingencies.
🎯 Exam Tip: To score full marks, construct each column step-by-step and write a clear concluding sentence stating whether the final column contains all T (tautology), all F (contradiction), or a mix of both (contingency).
Question 6. Using the truth table, verify:
(i) \( p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \)
(ii) \( p \rightarrow (p \rightarrow q) \equiv \sim q \rightarrow (p \rightarrow q) \)
(iii) \( \sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q) \equiv p \wedge q \)
Answer: By systematically evaluating each logical connective step-by-step, we can confidently establish the equivalence of these compound statements.
(i) Truth table for \( p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \):
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( r \) | \( q \wedge r \) | \( p \vee (q \wedge r) \) | \( p \vee q \) | \( p \vee r \) | \( (p \vee q) \wedge (p \vee r) \) |
| T | T | T | T | T | T | T | T |
| T | T | F | F | T | T | T | T |
| T | F | T | F | T | T | T | T |
| T | F | F | F | T | T | T | T |
| F | T | T | T | T | T | T | T |
| F | T | F | F | F | T | F | F |
| F | F | T | F | F | F | T | F |
| F | F | F | F | F | F | F | F |
The entries in columns 5 and 8 are identical.
\( \therefore p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \)
(ii) Truth table for \( p \rightarrow (p \rightarrow q) \equiv \sim q \rightarrow (p \rightarrow q) \):
| 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| \( p \) | \( q \) | \( \sim q \) | \( p \rightarrow q \) | \( p \rightarrow (p \rightarrow q) \) | \( \sim q \rightarrow (p \rightarrow q) \) |
| T | T | F | T | T | T |
| T | F | T | F | F | F |
| F | T | F | T | T | T |
| F | F | T | T | T | T |
The entries in columns 5 and 6 are identical.
\( \therefore p \rightarrow (p \rightarrow q) \equiv \sim q \rightarrow (p \rightarrow q) \)
(iii) Truth table for \( \sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q) \equiv p \wedge q \):
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( \sim q \) | \( p \rightarrow \sim q \) | \( \sim(p \rightarrow \sim q) \) | \( \sim(\sim q) \) | \( p \wedge \sim(\sim q) \) | \( p \wedge q \) |
| T | T | F | F | T | T | T | T |
| T | F | T | T | F | F | F | F |
| F | T | F | T | F | T | F | F |
| F | F | T | T | F | F | F | F |
The entries in columns 5, 7, and 8 are identical.
\( \therefore \sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q) \equiv p \wedge q \).
In simple words: A truth table helps us check if different logical expressions mean the exact same thing. By comparing the final columns for each expression, we can see that they have identical true and false values in every single row, which proves they are logically equivalent.
🎯 Exam Tip: Always number your columns clearly in truth tables and state which columns are identical to justify your final equivalence statement.
Question. Prove that \( \sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q) \equiv p \wedge q \) using a truth table.
Answer:
| \( 1 \) | \( 2 \) | \( 3 \) | \( 4 \) | \( 5 \) | \( 6 \) | \( 7 \) | \( 8 \) |
|---|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( \sim q \) | \( p \rightarrow \sim q \) | \( \sim(p \rightarrow \sim q) \) | \( \sim(\sim q) \) | \( p \wedge \sim(\sim q) \) | \( p \wedge q \) |
| T | T | F | F | T | T | T | T |
| T | F | T | T | F | F | F | F |
| F | T | F | T | F | T | F | F |
| F | F | T | T | F | F | F | F |
The entries in columns 5, 7 and 8 are identical.
\( \therefore \sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q) \equiv p \wedge q \).
In simple words: By building a truth table, we can see that the columns for \( \sim(p \rightarrow \sim q) \), \( p \wedge \sim(\sim q) \), and \( p \wedge q \) have the exact same true and false values in every row, which proves they are logically equivalent.
🎯 Exam Tip: Number your columns clearly in truth tables so you can easily refer to them in your final concluding statement to score full marks.
Question. Prove the following equivalence:
(iv) \( \sim(p \vee q) \vee (\sim p \wedge q) \equiv \sim p \)
Answer:
| \( 1 \) | \( 2 \) | \( 3 \) | \( 4 \) | \( 5 \) | \( 6 \) | \( 7 \) |
|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( \sim p \) | \( p \vee q \) | \( \sim(p \vee q) \) | \( \sim p \wedge q \) | \( \sim(p \vee q) \vee (\sim p \wedge q) \) |
| T | T | F | T | F | F | F |
| T | F | F | T | F | F | F |
| F | T | T | T | F | T | T |
| F | F | T | F | T | F | T |
The entries in columns 3 and 7 are identical.
\( \therefore \sim(p \vee q) \vee (\sim p \wedge q) \equiv \sim p \).
In simple words: We construct a truth table and compare the column for \( \sim p \) with the column for \( \sim(p \vee q) \vee (\sim p \wedge q) \). Since they match perfectly, the two statements are equivalent.
🎯 Exam Tip: Double-check the truth values for disjunction (\( \vee \)) and conjunction (\( \wedge \)) to avoid simple calculation errors in your table.
Question 7. Prove that the following pairs of statement patterns are equivalent:
(i) \( p \vee (q \wedge r) \) and \( (p \vee q) \wedge (p \vee r) \)
Answer:
| \( 1 \) | \( 2 \) | \( 3 \) | \( 4 \) | \( 5 \) | \( 6 \) | \( 7 \) | \( 8 \) |
|---|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( r \) | \( q \wedge r \) | \( p \vee (q \wedge r) \) | \( p \vee q \) | \( p \vee r \) | \( (p \vee q) \wedge (p \vee r) \) |
| T | T | T | T | T | T | T | T |
| T | T | F | F | T | T | T | T |
| T | F | T | F | T | T | T | T |
| T | F | F | F | T | T | T | T |
| F | T | T | T | T | T | T | T |
| F | T | F | F | F | T | F | F |
| F | F | T | F | F | F | T | F |
| F | F | F | F | F | F | F | F |
The entries in columns 5 and 8 are identical.
\( \therefore p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \).
In simple words: This is the distributive law of logical disjunction over conjunction. By comparing columns 5 and 8 in our 8-row truth table, we see they are identical, proving the equivalence.
🎯 Exam Tip: For three statements \( p, q, r \), your truth table must have exactly 8 rows to cover all possible combinations of truth values.
Question 1. Using truth tables, prove the following logical equivalences:
(i) \( p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \)
(ii) \( p \leftrightarrow q \equiv (p \rightarrow q) \wedge (q \rightarrow p) \)
(iii) \( p \rightarrow q \equiv \sim q \rightarrow \sim p \equiv \sim p \vee q \)
Answer:
(i) Truth table for \( p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \):
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( r \) | \( q \wedge r \) | \( p \vee (q \wedge r) \) | \( p \vee q \) | \( p \vee r \) | \( (p \vee q) \wedge (p \vee r) \) |
| T | T | T | T | T | T | T | T |
| T | T | F | F | T | T | T | T |
| T | F | T | F | T | T | T | T |
| T | F | F | F | T | T | T | T |
| F | T | T | T | T | T | T | T |
| F | T | F | F | F | T | F | F |
| F | F | T | F | F | F | T | F |
| F | F | F | F | F | F | F | F |
The entries in columns 5 and 8 are identical.
\( \implies p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) \)
(ii) Truth table for \( p \leftrightarrow q \equiv (p \rightarrow q) \wedge (q \rightarrow p) \):
| 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| \( p \) | \( q \) | \( p \leftrightarrow q \) | \( p \rightarrow q \) | \( q \rightarrow p \) | \( (p \rightarrow q) \wedge (q \rightarrow p) \) |
| T | T | T | T | T | T |
| T | F | F | F | T | F |
| F | T | F | T | F | F |
| F | F | T | T | T | T |
The entries in columns 3 and 6 are identical.
\( \implies p \leftrightarrow q \equiv (p \rightarrow q) \wedge (q \rightarrow p) \)
(iii) Truth table for \( p \rightarrow q \equiv \sim q \rightarrow \sim p \equiv \sim p \vee q \):
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \rightarrow q \) | \( \sim q \rightarrow \sim p \) | \( \sim p \vee q \) |
| T | T | F | F | T | T | T |
| T | F | F | T | F | F | F |
| F | T | T | F | T | T | T |
| F | F | T | T | T | T | T |
The entries in columns 5, 6, and 7 are identical.
\( \implies p \rightarrow q \equiv \sim q \rightarrow \sim p \equiv \sim p \vee q \)
In simple words: To prove that different logical statements are equivalent, we construct a truth table for each statement. If the columns representing the final outputs of these statements have identical truth values in every row, the statements are logically equivalent.
🎯 Exam Tip: Always number your columns clearly and state which columns are identical to justify your final equivalence statement. Double-check your truth values for conditional statements (\( \rightarrow \)) as they are only false when T leads to F.
Question (iii). Verify that \( p \rightarrow q \equiv \sim q \rightarrow \sim p \equiv \sim p \vee q \) using truth table.
Answer:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \rightarrow q \) | \( \sim q \rightarrow \sim p \) | \( \sim p \vee q \) |
| T | T | F | F | T | T | T |
| T | F | F | T | F | F | F |
| F | T | T | F | T | T | T |
| F | F | T | T | T | T | T |
The entries in columns 5, 6 and 7 are identical. This confirms that the conditional statement is logically equivalent to its contrapositive and its disjunctive form.
\( \therefore p \rightarrow q \equiv \sim q \rightarrow \sim p \equiv \sim p \vee q \).
In simple words: We can see that columns 5, 6, and 7 have the exact same pattern of True (T) and False (F) values. This means all three logical statements mean the exact same thing.
🎯 Exam Tip: Clearly number your columns in the truth table so you can easily refer to them in your final concluding statement to score full marks.
Question (iv). Verify the logical equivalence of \( \sim(p \wedge q) \) and \( \sim p \vee \sim q \) using truth table.
Answer:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|
| \( p \) | \( q \) | \( \sim p \) | \( \sim q \) | \( p \wedge q \) | \( \sim(p \wedge q) \) | \( \sim p \vee \sim q \) |
| T | T | F | F | T | F | F |
| T | F | F | T | F | T | T |
| F | T | T | F | F | T | T |
| F | F | T | T | F | T | T |
The entries in columns 6 and 7 are identical. This proves De Morgan's Law for the negation of a conjunction.
\( \therefore \sim(p \wedge q) \equiv \sim p \vee \sim q \).
In simple words: Since columns 6 and 7 have the exact same T and F values in every row, the two statements are logically identical. This shows that saying "not (A and B)" is the same as saying "not A or not B".
🎯 Exam Tip: Double-check your conjunction (\( \wedge \)) column before negating it, as a single error in column 5 will make your final columns mismatch.
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The complete and updated Maharashtra Board Class 12 Maths Part 1 Chapter 1 Mathematical Logic 1.6 Solutions is available for free on StudiesToday.com. These solutions for Class 12 Maths Commerce are as per latest MSBSHSE curriculum.
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