Could we build a truth table of "$P \Rightarrow \lnot P$"? What is its meaning and its Venn Diagram?

Solution 1:

Personally I like to think of $P \to Q$ as a "promise" or "contract":

I promise if $P$, then $Q$

So if I say:

I promise if "bob dies", then "I pay \$100"

• If bob dies ($P=T$), and don't pay \$100 ($Q=F$), I've broken my promise ($F$).
• If bob dies ($P=T$), and I pay \$100 ($Q=T$), I've kept my promise ($T$).
• If bob is alive ($P=F$), and I pay \$100 ($Q=F$), I've kept my promise ($T$).
  • Just because I paid you \$100, doesn't mean bobs dead.
• If bob is alive ($P=F$), and I don't pay \$100 ($Q=F$), I've kept my promise ($T$).

For $\lnot P \to P$

Your statement $\lnot P \to P$ can be interpreted as:

I promise if "I haven't paid \$100", then "I've paid \$100"

• If haven't paid \$100 ($P=F$), then I've broken the promise ($\lnot P \to P = F$).
• If I've paid \$100 ($P=T$), then I've kept the promise ($\lnot P \to P = T$)

From this we can deduce, the promise is broken only when I don't pay \$100. Or in other words it's all based on whether I've paid \$100 ($P$). Symbolically: $\lnot P \to P = P$

For $P \to \lnot P$

The statement $P \to \lnot P$ can be interpreted as:

I promise if "I have paid \$100", then "I haven't paid \$100"

• If haven't paid \$100 ($P=F$), then I've kept the promise ($\lnot P \to P = T$).
• If I've paid \$100 ($P=T$), then I've broken the promise ($\lnot P \to P = F$)

From this we can deduce, the promise is broken whenever I pay \$100. So it's the opposite of before ($\lnot P$). Symbolically: $P \to \lnot P = \lnot P$

As a venn diagram

A useful rule to keep in mind is that: $P \to Q = \lnot P \lor Q$.

So:

• $P \to \lnot P = \lnot P \lor \lnot P = \lnot P$
• $\lnot P \to P = \lnot \lnot P \lor P = P$

As a venn diagram this view also helps since: $\lnot P$ can be seen as "everything that's not in P".

And $\lor$ can be interpreted as combining the two venn-diagrams.

So your statement: $P \to \lnot P = \lnot P \lor \lnot P$, illustrated here:

Which you can see at the end just equals $\lnot P$.

For your other one: $\lnot P \to P = \lnot \lnot P \lor P = P \lor P$, illustrated here:

Which ends up as just $P$.

(I've written $u$ as the universe in the image out of habit, which you seem to have used $d$ instead, presumably as the domain of discourse which is also correct.)

Validity of $P \to \lnot P$

Usually when we talk about an argument being valid, we have a set of premises: {"Socrates is a man", "All men are mortal"}, and a conclusion "Socrates is mortal".

Symbolically: $(\text{"Socrates is a man"} \land \text{"All men are mortal"}) \to \text{"Socrates is mortal"}$

An equivalent definition of validity is:

An argument is valid if and only if it would be contradictory for the conclusion to be false if all of the premises are true

Here the premises is the set: {"Socrates is a man", "All men are mortal"}, and we ask:

Would it be contradictory for "Socrates is mortal" to be false if "Socrates is a man" and "All men are mortal" are both true?
Yes! Therefore it is valid.

If your asking if $P \to \lnot P$ is valid, we'll take the premises as the set $\{ P \}$, and we ask:

Would it be contradictory for $\lnot P$ to be false if $P$ was true?
No. This it is not contradictory, rather very natural.

So the statement $P \to \lnot P$ invalid.

In propositional logic, these valid arguments are tautologies, basically statements which are always true. As you can see $P \to \lnot P$ is not always true.

Solution 2:

$P \Rightarrow Q$ is like a claim.

Conventions vary, but some (many?) authors read $$P \Rightarrow Q$$ as a metalogical assertion that the logical operation (or truth function) $$P \to Q$$ is true (which means that either $P$ is false or $Q$ is true).

($P$ and $Q$ and $P \to Q$ and $P \Rightarrow Q$ are all sentences.)

$P$	$\lnot P$	$P \Rightarrow \lnot P$
True	False	False
False	True	True

Is an implication of the form $P \Rightarrow \lnot P$ valid or not valid?

The third column of the above truth table shows that the sentence $$P \to \lnot P$$ is a contingency, i.e., a sentence that is neither a tautology nor a contradiction. In propositional (though not in first-order) logic, tautology and validity are synonyms; so, the sentence is certainly not a validity.

Based on the above convention (where we read $\Rightarrow$ as asserting that $\to$ is true), the implication $$P \Rightarrow \lnot P$$ is therefore actually asserting that sentence $P$ is false.

But what is its meaning and venn diagram?

Since Venn diagrams deal with sets of objects, I'd rewrite $$P \Rightarrow \lnot P$$ as $$\forall x\;\big(P(x) \Rightarrow \lnot P(x)\big),$$ which is illustrated as a sunny-side-up where the yolk represents $P(x)$ and the entire egg represents $\lnot P(x).$ However, $P(x)$ and $\lnot P(x)$ are mutually exclusive and thus have an empty intersection. Hence, $P(x)$ is actually the empty set, i.e., $P(x)$ is true for no $x,$ i.e., $P$ is false. This conclusion is consistent with that in the previous paragraph.

Summarising: $$P \to \lnot P\quad \equiv\quad \lnot P.$$

Solution 3:

Not sure how useful a Venn diagram would be here, but if $P$ is true implies $P$ is false, then $P$ must be false.

Symbolically: $(P\implies \neg P)\implies \neg P$.

Here is the truth table:

Source: https://www.erpelstolz.at/gateway/TruthTable.html

Here is a formal proof by contradiction using a form of natural deduction (screenshot from my proof checker):

Solution 4:

I'd approach it via the adjunction $$R\to (S\to W)\ \ \iff\ \ (R\land S)\to W$$

Examples:

"If it's a rainy day, then if I'm also out on the street, then I'm gonna be wet."

"If it's a rainy day and I'm also out on the street, then I'm gonna be wet."

Now you can read $\neg P$ as $P\to \bot$ where $\bot$ denotes any false claim. To say $P$ is false is to say that it implies something absurd. With this,

$$P\to \neg P\ \ \iff\ \ P\to (P\to\bot)\ \ \iff\ \ (P\land P)\to \bot\ \ \iff\ \ \neg (P\land P)\iff\ \ \neg P$$

So $P\to \neg P$ is a weird way of saying $P$ is false, just like $\neg (P\land P)$ would be a weird way of saying it.

You could just tell your friend ($\neg P$)

"I would not enjoy having to wake up so early tomorrow."

Where I translate "$P$" to a the waking up task and the negation "$\bullet\to\bot$" to non-enjoyment.

But you could also tell your friend, along the lines of the latter variant ($\neg (P\land P)$)

"I would not enjoy, both having to wake up so early tomorrow and having to wake up so early tomorrow."

but the second is redundant from a classical logic perspective. Indeed so redundant that it would be weird to even say it.

Roughly framing would be along the lines of ($P\to \neg P$)

"If I would have to wake up so early tomorrow, then I would not enjoy it (having to wake up so early tomorrow)."

Indeed the semantics of $P\to (P\to Q)$ is that of $P\to Q$ for the same reason.