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I've encountered a textbook problem that's telling me to prove an iff statement or its negation.

Consider the statement:

$P \iff Q$

In my scenario, I have that:

$P \implies Q$ but it's not the case that $Q \implies P$

The negation of the original statement is $P \iff \neg Q$.

but clearly, this is not true because $P \implies Q$.

I am simply asking if my logic is correct, and both the statement and its negation are false.

user129393192
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  • According to the post I linked, the negation is no cloud <=> rain. The negation of $P \iff Q$ is $P \iff \neg Q$ – user129393192 Oct 29 '23 at 07:30
  • Your statement is equivalent to rain => clouds AND clouds => rain (and not to what you wrote). – bbbbbbbbb Oct 29 '23 at 07:30
  • @bbbbbbbbb what did I write I was stating the fact that rain => clouds is true and clouds => rain is not (real truth values). I'm saying the iff I'm facing is similar, where one way is True but the converse is not. – user129393192 Oct 29 '23 at 07:32
  • @Surb I am pretty sure what you say is the same statement. I'm referring to the accepted answer from post I linked. – user129393192 Oct 29 '23 at 07:34
  • @Surb This is not the negation. In fact those statements are equivalent. The negation of $A\iff B$ is $A\ xor B$. – Peter Oct 29 '23 at 07:37
  • That $A$ and $not\ A$ cannot be both true is what we call consistency. That one of $A$ and $not\ A$ must be true is what we call the law of the excluded middle. This law is generally accepted and important for many important results of mathematics. – Peter Oct 29 '23 at 07:41
  • @Peter The first thing you mention is the law of non-contradiction. Whereas consistency is about collections of axioms, or theories, I believe? – CM-rings Oct 29 '23 at 07:43
  • @freakish A statement cannot be equivalent to its negation. – Peter Oct 29 '23 at 07:45
  • @Peter can you explain what I am seeing? It seems like neither is true. – user129393192 Oct 29 '23 at 07:49
  • @CM-rings A system is called consistent if it is impossible that we can derive both $A$ and $not\ A$ for a statement $A$. I do not understand the comment. – Peter Oct 29 '23 at 07:49
  • @Peter The comment is that I think what you called consistency is actually called the law of non-contradiction. – CM-rings Oct 29 '23 at 07:52
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    Negation is not controversial at all: the negation of a true statement is a false one. – Mauro ALLEGRANZA Oct 29 '23 at 08:14
  • @MauroALLEGRANZA for my example, I am asking about… – user129393192 Oct 29 '23 at 08:24
  • In this case it seems the IFF statement and its negation are both false. Is this correct? That is my question. – user129393192 Oct 29 '23 at 08:24
  • No, an IFF is true when both have the same truth vue. Thus, it will be false when they have different truth values. – Mauro ALLEGRANZA Oct 29 '23 at 09:44
  • So the if and only if is false, but the negation is also false @MauroALLEGRANZA. Can you tell me about this? – user129393192 Oct 29 '23 at 18:26
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    Let us recap it: If P⟹Q is true and Q⟹P is false, then P⟺Q is false, simply because P⟺Q amounts to P⟹Q and Q⟹P. And yes, the negation of P⟺Q amounts to P⟺¬Q (check it with truth table). BUT this one is not the same as P⟹Q true and Q⟹P false, because the negation of P⟹Q is not P⟹¬Q. This is the reason why your conclusion does not hold. – Mauro ALLEGRANZA Oct 30 '23 at 13:59

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The statement $\ P\Rightarrow Q\ $ does not necessarily contradict $\ P\Leftrightarrow\neg\,Q\ $. You've specified that $\ Q\Rightarrow P\ $ is false, and this can be the case only when $\ P\ $ is false and $\ Q\ $ is true, and in that case both $\ P\Rightarrow Q\ $ and $\ P\Leftrightarrow\neg\,Q\ $ are true.

You need to keep in mind that the symbol $\ \Rightarrow\ $ represents material implication which has some properties that will appear counterintuitive if you confuse it with other forms of implication more commonly used outside formal logic. The proposition $\ P\Rightarrow R\ ,$ for instance, is always true whenever $\ P\ $ is false, regardless of what the proposition $\ R\ $ (or its truth value) is. In particular, both $\ P\Rightarrow Q\ $ and $\ P\Rightarrow\neg\,Q\ $ are true if and only if $\ P\ $ is false.

lonza leggiera
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  • Re: Material implication and common usage. Consider the implication $P \implies Q$. Suppose the antecedent $P$ is true. From the usual truth table, if the implication is true, then the consequent $Q$ must be true (the detachment rule). If the consequent $Q$ is false, then the implication must be false (the counter-example rule). These two rules sum up the common usage of implication. Rarely in daily discourse do we give much consideration to the implications of propositions known to be false. – Dan Christensen Oct 29 '23 at 16:01
  • ... In very technical arguments like mathematical proofs, however, we often consider such implications. Again, from the usual truth table for implication $P\implies Q$, if the antecedent $P$ is false, then the implication $P\implies Q$ will be true regardless of the truth value of the consequent $Q$ (the vacuous truth rule). – Dan Christensen Oct 29 '23 at 16:01
  • @DanChristensen the point is that implication is seen as stronger in natural usage. Most people don’t wake up, see that it’s cloudy and that it’s Tuesday, and then conclude “If it’s cloudy, then it’s Tuesday.” That is, the truth or falsehood of the propositional atoms of an implication are really only the truth-maker(s) for that implication in the context of Classical Logic. – PW_246 Oct 29 '23 at 16:19
  • @PW_246 It wouldn't be very useful, but they could rightly conclude that "if it's cloudy, then it's Friday." It might raise a few eyebrows, but it would not lead to any inconsistencies or errors in its acceptance as true at that instant in time. It makes no predictions about the future or claims about the past. – Dan Christensen Oct 29 '23 at 17:37
  • @DanChristensen yes, but that’s hardly what implication means in general. – PW_246 Oct 29 '23 at 18:58
  • @PW_246 Implication and causality are often confused (e.g. "smoking causes cancer" being equated to "smoking implies cancer"). Also, certain features of logical implication (e.g. vacuous truth) are rarely used in daily discourse, but they are apparently not inconsistent with more commonly used features (e.g. detachment and counter-examples). – Dan Christensen Oct 29 '23 at 20:05
  • @DanChristensen I never claimed that implication requires causality. The point is that at minimum, natural language conditionals are just not analyzable purely in terms of a case analysis of the truth or falsehood of its constituent propositions, but rather capture a notion similar to entailment, but without the need to be tied to a specific theory/logic/whatever. – PW_246 Oct 29 '23 at 20:36
  • AFAICT natural language conditionals involving true-or-false propositions in the present tense (or other fixed timeframe) are adequately modelled with material conditionals. Do you have a counter-example in mind? – Dan Christensen Oct 29 '23 at 20:54
  • Philosophers have been debating these issues. since at least as early as McColl (1908). I didn't intend to take any side in that debate. One suggestion, however, has been that "implication" in natural language means something closer to a form of entailment than to material implication—not necessarily logical or linguistic, but more in the sense that the consequent follows from the antecedent by a valid informal argument. – lonza leggiera Oct 30 '23 at 00:17
  • It therefore seemed to me that this might have been the reason for the OP's misunderstanding that $\ P\Rightarrow Q\ $ necessarily contradicted $\ P\Leftrightarrow\neg,Q\ .$ – lonza leggiera Oct 30 '23 at 00:18
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Consider the statement:

$P \iff Q$

So, you are assuming that: $~(P\implies Q)\land (Q\implies P)$

In my scenario, I have that:

$P \implies Q$ but it's not the case that $Q \implies P$

Now, you are assuming that: $~(P \implies Q) \land \neg (Q\implies P)$.

This leads to the obvious contradiction: $~(Q \implies P) \land \neg (Q\implies P)$

You can stop here.

Dan Christensen
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