Math Problem Statement

Construct truth table for (¬q ∧ (p → q)) → ¬p

Solution

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Problem 1: Reflexive and Symmetric Closures of R1 and R2

Given Data:

  • Set A={1,2,3,4}A = \{1, 2, 3, 4\}
  • Relation R1={(1,1),(1,2),(2,3),(3,4)}R1 = \{(1, 1), (1, 2), (2, 3), (3, 4)\}
  • Relation R2={(1,2),(2,2),(2,3),(3,2),(4,1),(4,4)}R2 = \{(1, 2), (2, 2), (2, 3), (3, 2), (4, 1), (4, 4)\}

Reflexive Closure:

A reflexive closure adds all the pairs (a,a)(a, a) where aAa \in A if they are not already in the relation.

  • For R1R1, the pairs already present are (1,1)(1, 1). We need to add (2,2)(2, 2), (3,3)(3, 3), and (4,4)(4, 4).

    • Reflexive closure of R1={(1,1),(1,2),(2,3),(3,4),(2,2),(3,3),(4,4)}R1 = \{(1, 1), (1, 2), (2, 3), (3, 4), (2, 2), (3, 3), (4, 4)\}
  • For R2R2, the pairs (2,2)(2, 2) and (4,4)(4, 4) are already present. We need to add (1,1)(1, 1) and (3,3)(3, 3).

    • Reflexive closure of R2={(1,2),(2,2),(2,3),(3,2),(4,1),(4,4),(1,1),(3,3)}R2 = \{(1, 2), (2, 2), (2, 3), (3, 2), (4, 1), (4, 4), (1, 1), (3, 3)\}

Symmetric Closure:

A symmetric closure adds (b,a)(b, a) for every (a,b)(a, b) in the relation if it is not already present.

  • For R1={(1,1),(1,2),(2,3),(3,4)}R1 = \{(1, 1), (1, 2), (2, 3), (3, 4)\}:

    • Add (2,1)(2, 1), (3,2)(3, 2), and (4,3)(4, 3).
    • Symmetric closure of R1={(1,1),(1,2),(2,3),(3,4),(2,1),(3,2),(4,3)}R1 = \{(1, 1), (1, 2), (2, 3), (3, 4), (2, 1), (3, 2), (4, 3)\}
  • For R2={(1,2),(2,2),(2,3),(3,2),(4,1),(4,4)}R2 = \{(1, 2), (2, 2), (2, 3), (3, 2), (4, 1), (4, 4)\}:

    • Add (2,1)(2, 1) and ((1, 4)\

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Math Problem Analysis

Mathematical Concepts

Logic
Truth Tables
Propositional Calculus

Formulas

p → q = ¬p ∨ q
¬(A ∧ B) = ¬A ∨ ¬B

Theorems

De Morgan's Laws
Implication Equivalence

Suitable Grade Level

Grades 10-12