Is $\neg q \equiv \neg q \lor p \equiv \neg q \lor \neg p$ a theorem?

Question

So I tried solving a theorem, and ended up with the following answer: $$\neg q \equiv \neg q \lor p \equiv \neg q \lor \neg p$$

Can a theorem be true even if proven with a negative variable? Doesn't $ \neg q$ render the theorem false?

I would have been pretty confident if I had: $$ q \equiv q \lor p \equiv q \lor \neg p$$ as an answer, because the theorem would be proven true because $q$ is true in all cases.

Thank you for your answer.. I might have made a mistake somewhere.

Answer 1

The statement

$$\neg q \equiv \neg q \lor p \equiv \neg q \lor \neg p$$

is a generalized biconditional with three terms. Generalized biconditionals are true if and only if an even number of its terms are false.

Now, if $q$ is true, then the first term $\neg q$ is definitely false, and of the other two terms $\neg q \lor p$ and $\neg q \lor \neg p$ exactly one will be true, since with $q$ being true, their truth-value solely depends on $p$. So, tow terms will be false, meaning that the statement is true.

On the other hand, if $q$ is false, then all three terms are true, so again the statement is true.

So, yes, the statement is a logical theorem.

Note that the same reasoning (just with the cases for $q$ being true and false flipped) would also show that your second statement

$$q \equiv q \lor p \equiv q \lor \neg p$$

is also a theorem.