Which statement is always true regardless of p and q?
Understand the Problem
The question is asking which of the provided logical statements is universally true regardless of the truth values of propositions p and q.
Answer
The correct answer is $p \lor \neg p$.
Answer for screen readers
The universally true statement is: $p \lor \neg p$.
Steps to Solve
- Analyze each statement for truth values
We need to evaluate each logical statement to determine if it is universally true regardless of the truth values of propositions $p$ and $q$.
- Evaluate statement a: $p \lor \neg p$
This statement represents a logical disjunction (OR) where one of the terms is the negation of $p$. This is a tautology because regardless of whether $p$ is true or false, the entire statement will always be true.
- Evaluate statement b: $\neg p \rightarrow p$
This is a conditional statement. If $p$ is true, then the statement is false (since $\neg p$ would be false). If $p$ is false, the statement is true. This is not universally true.
- Evaluate statement c: $p \rightarrow q$
This conditional statement states that if $p$ is true, then $q$ must also be true. However, this can be false if $p$ is true and $q$ is false. Thus, it is not universally true.
- Evaluate statement d: $p \land \neg p$
This statement represents a logical conjunction (AND) where one of the terms is the negation of $p$. This statement is never true since both $p$ and $\neg p$ cannot be true at the same time.
The universally true statement is: $p \lor \neg p$.
More Information
The statement $p \lor \neg p$ is known as the Law of Excluded Middle in propositional logic. It asserts that for any proposition $p$, either $p$ is true, or its negation $\neg p$ is true, making it a fundamental principle in classical logic.
Tips
Common mistakes include:
- Confusing the truth values of conditional statements. Remember that an implication $A \rightarrow B$ is only false when $A$ is true, and $B$ is false.
- Misinterpreting the conjunctions and disjunctions; for example, assuming $p \land \neg p$ can be true.