Math Problem Statement

Use a truth table to determine whether the statement is a​ tautology, a​ self-contradiction, or neither. left parenthesis p left right arrow q right parenthesis left right arrow left bracket left parenthesis q right arrow p right parenthesis logical and left parenthesis p right arrow q right parenthesis right bracket(p↔q)↔[(q→p)∧(p→q)] Question content area bottom Part 1 Complete the truth table. Part 2Part 3Part 4 p q p left right arrow qp↔q q right arrow pq→p p right arrow qp→q left parenthesis q right arrow p right parenthesis logical and left parenthesis p right arrow q right parenthesis(q→p)∧(p→q) left parenthesis p left right arrow q right parenthesis left right arrow left bracket left parenthesis q right arrow p right parenthesis logical and left parenthesis p right arrow q right parenthesis right bracket(p↔q)↔[(q→p)∧(p→q)] T T enter your response here enter your response here enter your response here enter your response here enter your response here T F enter your response here enter your response here enter your response here enter your response here enter your response here F T enter your response here enter your response here enter your response here enter your response here enter your response here F F enter your response here enter your response here enter your response here enter your response here enter your response here Part 5 Is the statement left parenthesis p left right arrow q right parenthesis left right arrow left bracket left parenthesis q right arrow p right parenthesis logical and left parenthesis p right arrow q right parenthesis right bracket(p↔q)↔[(q→p)∧(p→q)] a​ tautology, a​ self-contradiction, or​ neither?

Tautology

Neither

​Self-contradiction

Solution

We can use a truth table to analyze the statement:

(p↔q)↔[(q→p)∧(p→q)](p \leftrightarrow q) \leftrightarrow [(q \rightarrow p) \land (p \rightarrow q)]

Here's how we can approach it step by step.

Step 1: Truth table structure

We'll need to evaluate each component:

  1. pp
  2. qq
  3. p↔qp \leftrightarrow q (biconditional)
  4. q→pq \rightarrow p (implication)
  5. p→qp \rightarrow q (implication)
  6. (q→p)∧(p→q)(q \rightarrow p) \land (p \rightarrow q) (conjunction of implications)
  7. (p↔q)↔[(q→p)∧(p→q)](p \leftrightarrow q) \leftrightarrow [(q \rightarrow p) \land (p \rightarrow q)]

We'll evaluate the main logical expression in each case.

Truth Table

ppqqp↔qp \leftrightarrow qq→pq \rightarrow pp→qp \rightarrow q(q→p)∧(p→q)(q \rightarrow p) \land (p \rightarrow q)(p↔q)↔[(q→p)∧(p→q)](p \leftrightarrow q) \leftrightarrow [(q \rightarrow p) \land (p \rightarrow q)]
TTTTTTT
TFFTFFT
FTFFTFT
FFTTTTT

Step 2: Analyze the final column

The final column represents the overall truth value of the statement for each combination of truth values of pp and qq.

Step 3: Conclusion

Since the last column has all "T" (True) values, this statement is always true, which means the given statement is a tautology.

Answer: Tautology

Do you want further details on any specific step? Here are some related questions for practice:

  1. What is the definition of a biconditional statement p↔qp \leftrightarrow q?
  2. How does an implication p→qp \rightarrow q work in logic?
  3. Can you explain why (q→p)∧(p→q)(q \rightarrow p) \land (p \rightarrow q) is equivalent to p↔qp \leftrightarrow q?
  4. How do truth tables help in determining if a statement is a tautology or a contradiction?
  5. Could a compound statement involving biconditionals and implications ever be a contradiction?

Tip: In a biconditional p↔qp \leftrightarrow q, both pp and qq must have the same truth value for the statement to be true.

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Math Problem Analysis

Mathematical Concepts

Logic
Truth Tables
Tautology
Self-Contradiction

Formulas

(p ↔ q)
(q → p)
(p → q)
(q → p) ∧ (p → q)
(p ↔ q) ↔ [(q → p) ∧ (p → q)]

Theorems

Biconditional (↔)
Implication (→)
Conjunction (∧)

Suitable Grade Level

Grades 10-12

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