# Prove that these statements $(A \lor \neg B)$ and $\neg A \to \neg B$ are logically equivalent

I have these two statements -

$(A \lor \neg B)$ and $\neg A \to \neg B$

and i have to prove that they are logically equivalent, I don't know if I am doing my logic wrong though because I don't see how they are logically equivalent?

I did a truth table and it didn't work out as logically equivalent for me. Are they even logically equivalent or am I going wrong somewhere?

My truth table:

• Maybe you made a mistake in the truth table? – Alex Vong Dec 28 '17 at 15:14
• You say you did a truth table, so perhaps you should show us your work. – Lee Mosher Dec 28 '17 at 15:15
• If you can share your truth table work, we can probably help you find your error. The statements certainly are equivalent. "Pay, or you won't get served" is equivalent to "If you don't pay, then you won't get served". – G Tony Jacobs Dec 28 '17 at 15:16
• If $p$ is false then $p\to q$ is true (ex falso sequitur quodlibet). – drhab Dec 28 '17 at 15:31
• The first two entries in the right column are wrong. (F -> F is true, and F -> T is true) – The Chaz 2.0 Dec 28 '17 at 15:54

Take $A \vee \neg B$, we can see that with the disjunction the only way to get $0$ is to have $A \equiv 0$ and $B \equiv 1$, in all other cases it's $1$.
Take $\neg A \rightarrow \neg B$, we know that a implication is only $0$ if we have $1 \rightarrow 0$, but with the negations in our example, we can see that it's only $0$ if we have $0 \rightarrow 1$ (i.e. the negation of $1 \rightarrow 0)$
We can see that in both steps we get that the expression is only $0$ iff $A \equiv 0$ and $B \equiv 1$. This must then mean that they are logically equivalent (all other cases for both expressions become $1$ all around).