# How to negate if and only if

How can I get $$(A ↔ ¬B)$$ or $$(¬A ↔ B)$$ from $$¬(A ↔ B)$$ using propositional laws?

I tried expanding $$¬(A ↔ B)$$ and I got to $$(A ∧ ¬B) ∨ (B ∧ ¬A)$$

I also tried expanding $$(A ↔ ¬B)$$ and I got to $$(¬A ∨ ¬B) ∧ (B ∨ A)$$

I don't know where in my expansion I went off the wrong path

• If it helps, $\neg(A \Leftrightarrow B)$ is xor = "Exclusive or" = $A \oplus B$. en.wikipedia.org/wiki/Exclusive_or Nov 11, 2020 at 2:25
• I'm not sure I understand the question, but note that $\neg(A\leftrightarrow B)$ is symmetric in $A$ and $B$, and $A\leftrightarrow\neg B$ is not. So there's nothing you can do to the former to get to the latter. Nov 11, 2020 at 2:46
• As mark says you can get to them because that expression implies what you got, not necessarily what is in your first sentence. Nov 11, 2020 at 2:48

Your expansion of $$\lnot(A\leftrightarrow B)$$ into $$(A\land\lnot B)\lor(B\land\lnot A)$$ was correct.

Your expansion of $$(A\leftrightarrow\lnot B)$$ into $$(\lnot A\lor\lnot B)\land(B\lor A)$$ was also correct.

Well done. As @zkutch pointed out, the negation of $$A\leftrightarrow B$$ can also be though of as exclusive or (XOR), meaning one or the other but not both.

To address your comment, we first expand $$\lnot(A\leftrightarrow B)$$ into $$\lnot((A\land B)\lor(\lnot A\land\lnot B))$$. Then we use De Morgan's laws (which flip $$\land$$'s, $$\lor$$'s, and negate everything which isn't one) to get $$\lnot(A\land B)\land\lnot(\lnot A\land\lnot B)$$, which turns into $$(\lnot A\lor \lnot B)\land(A\lor B)$$.

We then distribute over the $$\land$$ to arrive at $$(\lnot A\land A)\lor (\lnot A\land B)\lor(\lnot B\land A)\lor(\lnot B\land B)$$, which simplifies to $$(\lnot A\land B)\lor(\lnot B\land A)$$.

Since $$(\lnot A\land B)\lor(\lnot B\land A)$$ reads as $$A$$ is true and not $$B$$, or $$B$$ is true and not $$A$$, it is equivalent to $$A\leftrightarrow\lnot B$$.

• But is there any way I can get from one to another? Think of it as a proving question and I want to show using proportional laws that ¬(A↔B) is equal to (A↔¬B). I've been thinking about this for a while now. Nov 11, 2020 at 2:34
• @UneducatedPotato Let me know if you get stuck at any point in the derivation. Nov 11, 2020 at 2:54
• Thanks for the explanation. I just wish there was a way we could show how (¬A∧B)∨(¬B∧A) is equivalent to A↔¬B @user400188 Nov 11, 2020 at 2:59
• @UneducatedPotato The very definition of $A$ if and only if $C$ is $(A\land C)\lor(\lnot A\land\lnot C)$. If you substitute $\lnot B$ for $C$, you will get your answer. Nov 11, 2020 at 3:25

Distribute:

$$(A\land\lnot B)\lor(B\land\lnot A)=(A\lor B)\land(\lnot B\lor B)\land(A\lor\lnot A)\land(\lnot B\lor\lnot A)$$

and then recognize that $$(P\lor\lnot P)$$ is always true, so the middle two pairs drop out and the right hand side reduces to

$$(A\lor B)\land(\lnot B\lor\lnot A)$$

• Ah, I didn't see that user400188 had edited their answer to include what amounts to the same thing. Nov 11, 2020 at 3:30