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I just saw that in the proof of $P\implies Q$, if P is $F$, then the implication is still $T$. This is so because when P is $F$, the implication is not getting disproven.

So when we find proof of $P \nRightarrow Q$, the answer I found is that the negation of implication is $T$ only when P is $T$ and Q is $F$. But I don't get one thing : When P is $F$, then how is $P \nRightarrow Q$ equal to $F$? Just like in previous case, if the antecedent is $F$ the implication is not disproven.

Similarly, if the antecedent here is $F$, then again the "does not imply" is not disproven, right?

So shouldn't $P \nRightarrow Q$ be $T$ when P is $F$ also?

Thank you for your time and patience!
This is my first question on mathematics exchange, so kindly point out any mistakes in my method of asking this question.
Regards

Prem
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Sahil Gupta
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  • Welcome to MSE. Can you please try to rewrite this question using appropriate logic symbols/operators to improve the readability? Doing this may help attract more comments and answers. – Red Five Apr 05 '24 at 07:08
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    Sorry! I have made the necessary edits. Kindly let me know if any other changes are usually required. Thank you – Sahil Gupta Apr 05 '24 at 07:20
  • Lovely. Thanks. – Red Five Apr 05 '24 at 07:30
  • The truth value of $P\implies Q$ is a definition , so cannot be proven. The explanation that it is true , if $P$ is false , is that in an "if"-statement , the "then"-part is irrelevant , if the "if"-part is false. – Peter Apr 05 '24 at 07:39
  • $\lnot (P \to Q)$ is erquivalent to $P \land \lnot Q$. – Mauro ALLEGRANZA Apr 05 '24 at 07:44
  • To prove that $P$ does not imply $Q$ you have to find a case (truth assignment, i.e. line in the truth table) where $P$ is True and $Q$ is False. – Mauro ALLEGRANZA Apr 05 '24 at 07:44
  • @Peter Though is often presented as The Definition, we can prove that $P \implies Q~~\equiv ~~ \neg (P \land \neg Q)$ from more fundamental rules of inference. See, for example, https://www.dcproof.com/DeriveImplies.html – Dan Christensen Apr 05 '24 at 15:53
  • "Just like in previous case, if the antecedent is F the implication is not disproven" There is not implication. There is a statement an impcation must fail. If the the antecedent is false the implication cant fail. The implication can only fail if the antecendent is true and the conclusion is false. In all other case the implication succeeds, of fails to fail. ... tl;dr $P\not\implies Q$ is just $\lnot (P\implies Q)$. $P\not\implies Q$ is true $\iff P\implies Q$ is false. Ad if $P$ is false $P\implies Q$ is true. So $P\not\implies Q$ MUST be false. – fleablood Apr 06 '24 at 03:39

3 Answers3

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The CORE ISSUE is about $\not \Rightarrow$ which should be avoided.

We should try to write $\lnot [ P \Rightarrow Q ]$ & not write $P \not \Rightarrow Q$ which is ambiguous.

Now , when $P$ is false , the Inner Implication $[ P \Rightarrow Q ]$ is true , since the Conclusion is not getting disproved , like you observed.

Then the Outer Negation automatically makes it true !

Basically , $[ P \Rightarrow Q ]$ & $\lnot [ P \Rightarrow Q ]$ ( which is improperly written like $P \not \Rightarrow Q$ ) are Negations of each other : Exactly 1 of them can be true while the other has to be false.

Prem
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  • I don't like $P \not\Rightarrow Q$ as a notation, but I don't see any possible ambiguity. Can you clarify your statement that it is ambiguous, please. – Rob Arthan Apr 06 '24 at 22:18
  • (1) OP here is giving a certain Interpretation to "P Does Not Imply Q" in the Sentence starting with "Similarly ..." , which is Basis for this Question Post. (2) This Question [[ https://math.stackexchange.com/questions/3906124/expressing-does-not-imply ]] gives one meaning to that , while the Accepted Answer there gives two meanings. [[Cont]] – Prem Apr 07 '24 at 08:53
  • [[Cont]] (3) This Question [[ https://math.stackexchange.com/questions/205388/not-necessarily-imply ]] & the Answers therein give more Interpretations. (4) I have 3 additional Interpretations for this. It is too cumbersome to list those here since the nuances are very fine , while the Example Statements will be very long. (5) Since the ambiguity is already apparent with the other Questions , @RobArthan , I will not go into the Details of more Interpretations. (6) My Observation : Ambiguity is in the eye of the beholder. – Prem Apr 07 '24 at 09:02
  • please don't say something is to be avoided or ambiguous without giving concrete reasons. Mathematics is a science and mathematical assertions should be backed up by evidence. – Rob Arthan Apr 07 '24 at 19:10
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Let $P$ be "it rains" and $Q$ be "the road is wet".

I'm using the $\neg$ symbol to notate negation, so if $P=T$, $\neg P = F$.

Now if the road has no roof, clearly $P \Rightarrow Q = T$.
Also if it doesn't rain, the road may remain dry $\neg P \Rightarrow \neg Q = T$, or it can get wet by other means $\neg P \Rightarrow Q = T$.

Now let's look at $\neg(\neg P \Rightarrow Q)$. According to our analysis above, $\neg P \Rightarrow Q = T$. So $\neg(\neg P \Rightarrow Q) \neg T = F$.

In the words of our example:

$\neg P \Rightarrow Q$ means even though it's not raining, the road can still get wet. Negating that statement means "even though it's raining, the road remains dry", which doesn't work, since we already established there's no roof.

This is of course what @Mauro ALLEGRANZA said in the comments:

$\neg(P \Rightarrow Q) = P \text{ and } \neg Q$.

infinitezero
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In classical propositional logic, by saying that "P implies Q", we are simply ruling out the possibility that both P is true and Q is false.

$P \implies Q~~\equiv ~~\neg (P \land \neg Q)$

To answer your question, negating both sides, we have:

$\neg(P \implies Q)~~\equiv ~~P \land \neg Q$

"Real-world" Example

What do we mean, in propositional logic, by "if it is raining, then it is cloudy?"

We do not mean that rain causes cloudiness.

We also do not mean that whenever it is raining, it is always cloudy, a counter-example being a so-called sunshower.

We are simply ruling out that, at present, it is both raining and not cloudy.

$Raining \to Cloudy~~\equiv ~~\neg (Raining \land \neg Cloudy)$

Dan Christensen
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  • +1: the key point is that material implication is not to do with causality: causality is outside the scope of classical propositional and predicate logic. – Rob Arthan Apr 06 '24 at 22:21