Truth Table Analysis for (p ∨ q) → ¬ (¬p ∧ ¬q) - Tautology, Contingency, or Contradiction?

Math Problem Statement

Construya la tabla de verdad de la fórmula lógica: (p ∨ q) → ¬ (¬p ∧ ¬q). Indique si es tautológica, contingente o contradictoria.

To solve this problem, we need to build a truth table for the formula:

$(p \vee q) \rightarrow \neg (\neg p \wedge \neg q)$

Identify the Components:
- $p$ and $q$ : Propositional variables.
- $p \vee q$ : "p OR q."
- $\neg p$ : NOT $p$ .
- $\neg q$ : NOT $q$ .
- $\neg p \wedge \neg q$ : "NOT $p$ AND NOT $q$ " (conjunction of negations).
- $\neg (\neg p \wedge \neg q)$ : Negation of the conjunction above.
- $(p \vee q) \rightarrow \neg (\neg p \wedge \neg q)$ : Final expression, which is an implication.
Truth Table: Construct a truth table to evaluate the truth values for each component based on all possible values of $p$ and $q$ .

\hline p & q & p \vee q & \neg p & \neg q & \neg p \wedge \neg q & \neg (\neg p \wedge \neg q) & (p \vee q) \rightarrow \neg (\neg p \wedge \neg q) \\ \hline T & T & T & F & F & F & T & T \\ T & F & T & F & T & F & T & T \\ F & T & T & T & F & F & T & T \\ F & F & F & T & T & T & F & T \\ \hline \end{array}$$ ### Analysis: - The final column $$(p \vee q) \rightarrow \neg (\neg p \wedge \neg q)$$ is **always true** regardless of the values of $$p$$ and $$q$$. ### Conclusion: Since the formula $$(p \vee q) \rightarrow \neg (\neg p \wedge \neg q)$$ is true for all possible truth values of $$p$$ and $$q$$, it is a **tautology**. Would you like further details or have any questions about the steps? --- Here are some related questions to explore this topic: 1. What defines a statement as a tautology, contradiction, or contingency? 2. How does the truth table for an implication work, specifically when the antecedent is false? 3. Can you explain why $$(p \vee q) \rightarrow \neg (\neg p \wedge \neg q)$$ is always true using logical laws? 4. How do De Morgan’s laws apply to simplify $$\neg (\neg p \wedge \neg q)$$? 5. What is the difference between an implication and a biconditional statement in logic? **Tip**: Remember, in a truth table, the implication $$A \rightarrow B$$ is true if $$A$$ is false, regardless of $$B$$'s truth value.

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Propositional Logic
Truth Tables
Logical Formulas

(p ∨ q) → ¬ (¬p ∧ ¬q)

Implication in Logic
De Morgan's Laws

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