It Makes Sense If You Understand Decision Theory – a podcast literary analysis of Planecrash/Project Lawful
Episode 5 – Content Warning: Math
For Next Week, reading goes to: “nor, indeed, the same book, same language, same library, same city, same planet, or same laws of physics as Golarion.”
The Official IMSIYUDT Weekly Reading Table of Contents, created and maintained by Cakoluchiam!
The app to render it to phones (you gotta change the Week_X number yourself)
Epub versions of Planecrash
You can support this podcast at our Patreon. We have a Discord as well.
Podcast: Play in new window | Download | Embed
Subscribe: RSS
RSS – All Posts
iTunes
Stitcher Smart Radio
Google Play
Email – Episodes only
Yes, you are misunderstanding the math. There is no typo.
Let’s consider the claim, “If someone liked HPMoR, then they will like Project Lawful.”
Is this claim true or false? To show that it is false, you should be able to find a counterexample. What do I mean by a counterexample? I mean somebody who liked HPMoR but didn’t like Project Lawful. If you can find such a person, then the claim is false.
But nobody who DIDN’T like HPMoR could possibly serve as a counterexample to my sentence. The claim doesn’t say anything about those people. If you find someone who didn’t like HPMoR and didn’t like Project Lawful, that’s not contrary to my claim. If you find someone who didn’t like HPMoR and DID like Project Lawful, that’s not contrary to my claim either.
This is a specific example to point out a general principle that, for any statement of the form “If [premise], then [conclusion]”, the statement is trivially true in cases when the premise is false. The statement made no claims whatsoever about what should be true if the premise is false, so no possible outcome contradicts the statement in that case.
Some examples:
“If a shape is a square, then it is a quadrilateral” is true of all shapes. Triangles are not counterexamples, since the statement makes no claims about non-squares.
“If a number is prime, then it is odd” is false. But the only counterexample is “2” (which IS prime). “6” or “9” do not constitute counterexamples, since the statement makes no claims about non-primes.
All sentences of the form “If someone is naked, then X” are true of clothed people. The only possible counterexample is someone who is naked but not X; the sentence makes no claims about nonnaked people.
This is true even if we fill in X with “they are wearing a shirt”, such that the full sentence becomes “If someone is naked, then they are wearing a shirt.” This sentence is true of people wearing shirts, since it makes no claims about people wearing shirts. Of course, we can easily disprove the sentence by pointing to a naked person. The point isn’t that the statement is necessarily true, just that people who have shirts on aren’t the reason it isn’t. “If someone IN THIS ROOM is naked, then they are wearing a shirt” is in fact true so long as no one in the room is naked.
More symbolically, we can write a Truth Table for p → q, where we consider each of the 4 possible combinations of truth values for p and q and write what the truth value of p → q is in that case. I’m using it to summarize what I said in words above, but this is also one possible way of formally defining what we mean by a logical operator like “→” in the first place.
p&em;&em;q&em;&em;p→q
T&em;&em;T&em;&em;T
T&em;&em;F&em;&em;F
F&em;&em;T&em;&em;T
F&em;&em;F&em;&em;T
We have written here in chart form the same thing that was said above in words; namely, p → q is interestingly true when q is true , but also trivially true when p is false. The only row in which p → q is false is when p is true and q is false.
We can keep going with this approach to solve the ((p → q) → p) → p puzzle. If we apply the “→” rule again to the p → q column and the p column, continuing to keep in mind that the only case that should produce an F is if the left thing is T but the right thing is F, we get:
p&em;&em;q&em;&em;p→q&em;(p→q)→p
T&em;&em;T&em;&em;T&em;&em;T
T&em;&em;F&em;&em;F&em;&em;T
F&em;&em;T&em;&em;T&em;&em;F
F&em;&em;F&em;&em;T&em;&em;F
To phrase it the same way I did last time, (p → q) → p is interestingly true when p → q and p are true, but also trivially true when p → q is false. The rows where (p → q) → p is false are when p → q is true but p is false.
But notice that the (p → q) → p column is identical to the p column. We have just proven using the truth tables that (p → q) → p = p. In words, (p → q) → p means “p → q is not trivially true”, which is the same as “p is not false”, which is the same as “p”.
Therefore, the last step is super easy:
p&em;&em;q&em;&em;p→q&em;(p→q)→p&em;((p→q)→p)→p
T&em;&em;T&em;&em;T&em;&em;T&em;&em;T
T&em;&em;F&em;&em;F&em;&em;T&em;&em;T
F&em;&em;T&em;&em;T&em;&em;F&em;&em;T
F&em;&em;F&em;&em;T&em;&em;F&em;&em;T
That is, p → p is always true.
More symbolically, we can write a Truth Table for p → q, where we consider each of the 4 possible combinations of truth values for p and q and write what the truth value of p → q is in that case. I’m using it to summarize what I said in words above, but this is also one possible way of formally defining what we mean by a logical operator like “→” in the first place.
p q p→q
T T T
T F F
F T T
F F T
We have written here in chart form the same thing that was said above in words; namely, p → q is interestingly true when q is true , but also trivially true when p is false. The only row in which p → q is false is when p is true and q is false.
We can keep going with this approach to solve the ((p → q) → p) → p puzzle. If we apply the “→” rule again to the p → q column and the p column, continuing to keep in mind that the only case that should produce an F is if the left thing is T but the right thing is F, we get:
p q p→q (p→q)→p
T T T T
T F F T
F T T F
F F T F
To phrase it the same way I did last time, (p → q) → p is interestingly true when p → q and p are true, but also trivially true when p → q is false. The rows where (p → q) → p is false are when p → q is true but p is false.
But notice that the (p → q) → p column is identical to the p column. We have just proven using the truth tables that (p → q) → p = p. In words, (p → q) → p means “p → q is not trivially true”, which is the same as “p is not false”, which is the same as “p”.
Therefore, the last step is super easy:
p q p→q (p→q)→p ((p→q)→p)→p
T T T T T
T F F T T
F T T F T
F F T F T
That is, p → p is always true.
I’m listening to these with some delay, so just got to it, but I really enjoyed the lecture highlights here.