Yes, the logical implication $p \to q$ (read as "if $p$, then $q$") is indeed logically equivalent to $\neg p \lor q$ (read as "not $p$ or $q$"). This is a well-known equivalence in propositional logic.

Here’s why:

• $p \to q$ can be interpreted as saying that "if $p$ is true, then $q$ must also be true."
• This is equivalent to saying that either $p$ is false (i.e., $\neg p$ is true), or $q$ is true.

Hence, $p \to q \equiv \neg p \lor q$.

Truth table for verification:

$p$	$q$	$p \to q$	$\neg p \lor q$
T	T	T	T
T	F	F	F
F	T	T	T
F	F	T	T

Both expressions yield the same truth values, proving they are logically equivalent.

Would you like to explore more on logical equivalences or anything else?

Related Questions:

1. What is the truth table for $p \lor q$?
2. Can you explain the difference between $p \to q$ and $p \leftrightarrow q$?
3. How do De Morgan's laws relate to $\neg p \lor q$?
4. What are tautologies and contradictions in logic?
5. How can we prove $p \to q \equiv \neg p \lor q$ using Boolean algebra?

Tip:

Always use truth tables when trying to establish logical equivalences—they visually demonstrate the consistency of two logical statements!