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TRUTH TABLE | LOGIC HSC | 12TH STANDARD MATHS | Mathematical Logic.

Mathematical Logic | Truth Tables | HSC 12th Standard Maths

Mathematical Logic: Truth Tables & Solutions (HSC 12th)

HSC 12th Board Exam Papers

Solutions and Explanations

Below are the detailed explanations and solutions for Mathematical Logic, specifically focusing on logical connectives and the construction of truth tables, which is a core part of the HSC 12th Standard Mathematics syllabus.

1. Introduction to Logical Connectives

A simple statement is a declarative sentence which is either true or false, but not both simultaneously. We use logical connectives to join simple statements to form compound statements.

Key Symbols:
  • Conjunction (AND): \( \wedge \)
  • Disjunction (OR): \( \vee \)
  • Negation (NOT): \( \sim \)
  • Conditional (Implication): \( \rightarrow \)
  • Biconditional (Double Implication): \( \leftrightarrow \)

2. Fundamental Truth Tables

Before solving complex problems, we must understand the standard truth values for each connective.

A. Conjunction (\( p \wedge q \)) and Disjunction (\( p \vee q \))

p q p \( \wedge \) q (AND) p \( \vee \) q (OR)
T T T T
T F F T
F T F T
F F F F

Note: Conjunction is True only if both are True. Disjunction is False only if both are False.

B. Conditional (\( p \rightarrow q \)) and Biconditional (\( p \leftrightarrow q \))

p q p \( \rightarrow \) q p \( \leftrightarrow \) q
T T T T
T F F F
F T T F
F F T T

Note: Implication is False only when Hypothesis (p) is True and Conclusion (q) is False. Biconditional is True when both have the same truth value.


Solved Examples: Constructing Truth Tables

Question 1: Construct the truth table for \( (p \wedge q) \vee \sim p \)

Solution:

We need columns for \( p \), \( q \), \( \sim p \), \( p \wedge q \), and finally the whole statement.

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

Question 2: Examine whether the statement pattern is a Tautology, Contradiction, or Contingency.

Statement: \( (p \rightarrow q) \leftrightarrow (\sim p \vee q) \)

Solution:

A Tautology is true in all cases. A Contradiction is false in all cases. A Contingency is a mix.

p q \( \sim p \) \( p \rightarrow q \) (I) \( \sim p \vee q \) (II) (I) \( \leftrightarrow \) (II)
T T F T T T
T F F F F T
F T T T T T
F F T T T T
Conclusion: Since all the entries in the last column are 'T', the given statement pattern is a Tautology. This also proves that \( p \rightarrow q \) is logically equivalent to \( \sim p \vee q \).

Question 3: Three variable Truth Table

Construct the truth table for: \( (p \vee q) \rightarrow r \)

Solution:

Since there are 3 statements (p, q, r), there will be \( 2^3 = 8 \) rows.

p q r p \( \vee \) q \( (p \vee q) \rightarrow r \)
T T T T T
T T F T F
T F T T T
T F F T F
F T T T T
F T F T F
F F T F T
F F F F T

Important Rules to Remember (Summary)

  • Negation: \( \sim T = F \) and \( \sim F = T \)
  • Conjunction: T only if T \( \wedge \) T
  • Disjunction: F only if F \( \vee \) F
  • Conditional: F only if T \( \rightarrow \) F
  • Biconditional: T if values match (T \( \leftrightarrow \) T or F \( \leftrightarrow \) F)