Math Problem Statement

~q → (p ∨ ~r) truth table

Solution

To create a truth table for the expression q(pr)\sim q \to (p \vee \sim r), we first rewrite the implication ABA \to B using its equivalent form: AB\sim A \vee B. Thus, the given expression can be rewritten as:

q(pr)(q)(pr)q(pr).\sim q \to (p \vee \sim r) \equiv \sim (\sim q) \vee (p \vee \sim r) \equiv q \vee (p \vee \sim r).

This means we need to evaluate the expression q(pr)q \vee (p \vee \sim r). Let's construct the truth table step by step:

Variables

  • pp
  • qq
  • rr

Derived Columns

  1. q\sim q
  2. r\sim r
  3. prp \vee \sim r
  4. q(pr)\sim q \to (p \vee \sim r)

Truth Table

ppqqrrq\sim qr\sim rprp \vee \sim rq(pr)\sim q \to (p \vee \sim r)
TTTFFTT
TTFFTTT
TFTTFTT
TFFTTTT
FTTFFFT
FTFFTTT
FFTTFFF
FFFTTTT

Explanation

  1. Column q\sim q: The negation of qq.
  2. Column r\sim r: The negation of rr.
  3. Column prp \vee \sim r: The logical OR between pp and r\sim r.
  4. Column q(pr)\sim q \to (p \vee \sim r): Evaluated as q(pr)\sim q \vee (p \vee \sim r), combining the results.

Let me know if you'd like more detailed steps for each column! 😊


5 Related Questions

  1. What is the difference between logical implication (\to) and equivalence (\leftrightarrow)?
  2. How do truth tables help verify logical equivalences?
  3. Can we derive pqp \vee q from qp\sim q \to p? Why or why not?
  4. What is the purpose of using De Morgan's Laws in logical expressions?
  5. How does (p)\sim (\sim p) simplify in logical expressions?

Tip

When constructing a truth table, ensure each variable has 2n2^n rows, where nn is the number of distinct variables. This ensures all combinations are covered!

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Math Problem Analysis

Mathematical Concepts

Logic
Truth Tables
Logical Implication
Logical Negation
Logical Disjunction

Formulas

A → B ≡ ~A ∨ B
~q → (p ∨ ~r) ≡ ~(~q) ∨ (p ∨ ~r)

Theorems

Implication equivalence: A → B ≡ ~A ∨ B
De Morgan's Laws

Suitable Grade Level

Grades 10-12