# Is the negation of "p iff q", "(not p) iff q"?

My logic teacher says that the negation of $$p \!\!\iff\!\! q$$ is $$(\lnot p) \!\!\iff\!\! q$$. This seems wrong to me, because I feel like $$(\lnot p) \!\iff\! q$$ is a too strong statement to be the logical consequence of $$\lnot (p \!\iff\! q)$$.

To illustrate to me that $$(\lnot p) \!\iff\! q$$ is the negation of $$p \!\iff\! q$$, my professor showed me a truth table that showed that $$(\lnot p) \!\iff\! q$$ is true iff $$p \!\iff\! q$$ is false.

Is my professor right? Can someone explain to me.

An example to illustrate why this statement concerns me:

Let $$p$$ be "It is Thursday"

Let $$q$$ be "It is raining"

$$p \!\iff\! q$$ is not true because there exists days where it is Thursday and it is not raining, yet $$(\lnot p) \!\iff\! q$$ is also not true because there are days when its Thursday and it does rain.

• Truth tables never lie (as long as they're correct). They are not always insightful, though. Sep 21 '18 at 16:08
• @Arthur can you explain to me what's wrong with my line of thinking in the example I provided? (Just edited my original post) thanks. Sep 21 '18 at 16:12
• Here is the truth table wolframalpha.com/input/… Sep 21 '18 at 19:21

p iff q is not true because there exists days where it is Thursday and it is not raining, yet (not p) iff q is also not true because there are days when its Thursday and it does rain.

You are missing a quantifier. You are comparing a statement of the form "Everyday X holds" to "Everyday not X holds". There is no reason those should be opposites. It is possible for those both to be false, just allow for some days X and some days not X.

The proper comparison would be

"Every day it is Thursday iff it is raining" is false

vs

"Some days it is not thursday iff it is raining" is ????

Work out what goes into ???? and I think you'll have it.

Let's work this out using only laws of equivalence, so you see that the negation of $$p \leftrightarrow q$$, is equivalent to $$\lnot p \leftrightarrow q$$.

\begin{align} \lnot(p \leftrightarrow q) &\equiv \lnot\Big((p \to q) \land (q\to p)\Big)\tag{Def. of \leftrightarrow}\\ \\ &\equiv \lnot \Big((\lnot p \lor q) \land (\lnot q \lor p)\Big)\tag{Def. of implication} \\ \\ &\equiv \lnot(\lnot p \lor q) \lor \lnot(\lnot q\lor p)\tag{DeMorgan's}\\ \\ &\equiv (p \land \lnot q)\lor (q \land \lnot p)\tag{DeMorgan's} \\ \\ &\equiv (p\lor q) \land (p \lor \lnot p) \land (\lnot q \lor q) \land (\lnot q \lor \lnot p)\tag{Distributivity}\\ \\ &\equiv (p\lor q)\land (\lnot q \lor \lnot p)\\ \\ &\equiv (\lnot p \to q) \land (q \to \lnot p)\tag{Def. of implication}\\ \\ &\equiv (\lnot p)\leftrightarrow q\tag{Def. of biconditional} \end{align}

$$P\iff Q$$ is the same as $$(P\implies Q)\text { and } (Q\implies P)$$

Therefore the negation is $$(P \text { and not } Q) \text { or } (Q\text { and not } P)$$

• I completely understand that, and I agree. However, my professor says it is also true that the negation is (not P) ⟺ Q Is this true? And if not, can you explain? Sep 21 '18 at 16:16
• Your teacher is correct. If you check the truth table for (not P) ⟺ Q it will be the same as for not ( P ⟺ Q) Sep 21 '18 at 17:14

With respect to your example, what you are talking about is actually about predicate logic, or maybe even modal logic, more advanced topics that I will not enter in here. Hence, your example is not exactly right.

As someone said in the comments, truth tables are always right, so your professor is right. However maybe the following example will be a little bit more insightful:

$$P \iff Q$$: $$0$$ is the least integer IFF every other integer is greater than $$0$$

This is, of course true, hence the negation, according to your professor:

$$\lnot P \iff Q$$: $$0$$ is not the least integer IFF every other integer is greater than $$0$$

Which is of course false, so, at least for this example, your proffessor is intuitively right.

Perhaps a more intutive way of grasping what IFF mean, is the following: P IFF Q, means that whenever happens P, also happens Q, and viceversa. So it's natural to think of the negation as: Whenever P doesn't happen, Q happens, and viceversa; that is, $$\lnot$$P IFF Q.

Let p be "It is Thursday"

Let q be "It is raining"

p iff q is not true because there exists days where it is Thursday and it is not raining, yet (not p) iff q is also not true because there are days when its Thursday and it does rain.

You're thinking about $$p(t)$$ being the statement "at time $$t$$ it's Thursday" and $$q(t)$$ being "at time $$t$$ it is raining", and then comparing $$\forall t(p(t)\iff q(t))\quad \text{to}\quad\forall t(\lnot p(t)\iff q(t))$$ and these are clearly both untrue, as you rightly point out. However, this is to be expected, because $$\forall s(r(s))$$ and $$\forall s(\lnot r(s))$$ for some statement $$r$$ are not inverses in general.

You will find that for any fixed point $$t$$ in time, $$p(t)\iff q(t)$$ and $$\lnot p(t)\iff q(t)$$ are inverses.

First, the two formulas $$\lnot (p \leftrightarrow q)$$ and $$(\lnot p) \leftrightarrow q$$ are logically equivalent because they have the same truth table, so they have the same meaning (at least, from a logical point of view). Said differently, $$(\lnot p) \leftrightarrow q$$ is the negation of $$p \leftrightarrow q$$.

In your example, you claim that $$p\leftrightarrow q$$ is not true because there exists a day where it is Thursday (i.e. $$p$$ is true) and it is not raining (i.e. $$q$$ is false). But then, in such a day $$(\lnot p) \leftrightarrow q$$ is true because $$\lnot p$$ and $$q$$ are both false. More generally, you can easily convince yourself that, for every day, $$p \leftrightarrow q$$ is true as soon as $$(\lnot p) \leftrightarrow q$$ is false, and vice-versa: you should check just 4 possibilities (it is Thursday or not, it is raining or not). So, even in your example $$(\lnot p) \leftrightarrow q$$ is the negation of $$p \leftrightarrow q$$.

Intuitively, in your example you have "cheated" because you compare the sentences $$p \leftrightarrow q$$ and $$(\lnot p) \leftrightarrow q$$ when they refer to distinct days. This means that $$p$$ in the first sentence $$p \leftrightarrow q$$ has a different meaning from $$p$$ in the second sentence $$(\lnot p) \leftrightarrow q$$.

• @amWhy - Sorry, I don't understand the meaning of your edit. Sep 21 '18 at 22:51
• I just tried to abbreviate the first sentence. I'm not sure what I did to get the emphasized text. Apologies. I've deleted that. Sep 21 '18 at 22:54
• @amWhy - Ah, that's OK, thank you. Sep 21 '18 at 23:08