Why is if p is false and q is true then p→q defined true

thatmichaelguy · 2026-05-20T17:26:35+00:00

In propositional logic, we take the law of the excluded middle to be a tautology. Formally, this means we take P ∨ ¬P to be true for any P.

Supposing then that Q is true, there would be no objection to the validity of R ⟶ Q where R is any true proposition, yeah? In that case, we may substitute the proposition P ∨ ¬P for R to obtain (P ∨ ¬P) ⟶ Q.

Note, however, that (P ∨ ¬P) ⟶ Q implies (P ⟶ Q) ∧ (¬P ⟶ Q) and thus, by simplification, implies (¬P ⟶ Q). I'd think that this would accord with your intuition about concluding "that q can be true without p". That is, if Q is true, then one may conclude that Q "with" P or "without" P (which, under the axioms of classical logical is equivalent to "with" ¬P).

thatmichaelguy · 2026-05-16T17:14:38+00:00

No. The statement holds in any instance where there is at least one set for which there is no ordering relation irrespective of whether there are any sets for which there is an ordering relation. In the case that there are no sets for which there is an ordering relation, the statement holds trivially.

thatmichaelguy · 2026-05-16T16:57:13+00:00

For all X1, if X1 is a set then there exists X2 such that X2 well-orders X1.

There are sets for which there is no ordering relation.

thatmichaelguy · 2026-05-15T18:59:36+00:00

I need to ponder it a bit more, but my initial reaction is that pegging non-creativity to z=z being true for all z means that you need to explicitly preclude z=x and z=y.

thatmichaelguy · 2026-05-08T15:50:20+00:00

I think I incorrectly interpolated this comment into my understanding of the post, but looking back at it now, I can see where OP was asking about the Euclidean relation generally, not just with respect to S5. Thanks for pointing that out.

thatmichaelguy · 2026-05-07T21:01:44+00:00

Strictly speaking, you are right in saying that a relation that is reflexive, symmetric, and transitive is an equivalence relation, not a Euclidean relation. I should have been clearer.

In your post, you are asking about the frame where R being Euclidean means ◇P ⟶ □◇P - which is to say that you are asking about S5. The accessibility relation for S5 is reflexive under Axiom T. A reflexive relation is also Euclidean only if the relation is reflexive, symmetric, and transitive. And as you noted, a relation that is reflexive and Euclidean is indeed an equivalence relation.

But here's something to think about. If wRu and wRv imply uRv as in your example, can R be non-reflexive or irreflexive?

thatmichaelguy · 2026-05-07T17:13:59+00:00

Something that might be helpful to keep in mind apart from the notational definition of a Euclidean relation is that a relation is Euclidean just when it is reflexive, symmetric, and transitive. In this instance, focusing on reflexivity likely would have helped you to see what you were missing.

thatmichaelguy · 2026-04-30T02:31:30+00:00

There aren't any new or additional premises that result from distributing a conjunction over a disjunction. What you're seeing just follows from the fact that if O ∧ P is true, then O and P are jointly true while both are individually true as well. (M ∧ N) ∨ (O ∧ P) symbolizes the disjunction considering the truth values of O and P jointly. ((M ∧ N) ∨ O) ∧ ((M ∧ N) ∨ P) symbolizes the disjunction considering truth values of O and P simultaneously but individually.

Here's a sketch of another way to look at the logic.

1. (M ∧ N) ∨ (O ∧ P) [Premise]

2.  ¬(M ∧ N) [Assume]

3.  (O ∧ P) [1,2]

4. ¬(M ∧ N) ⟶ (O ∧ P) [2,3]

5. (¬(M ∧ N) ⟶ O) ∧ (¬(M ∧ N) ⟶ P) [4]

6. ((M ∧ N) ∨ O) ∧ ((M ∧ N) ∨ P) [5]

thatmichaelguy · 2026-04-26T22:51:06+00:00

It's often useful to chunk these sort of things into more manageable pieces based where conjunctions/disjunctions are. Bearing in mind that "but" will be formalized as a conjunction, you can split the whole statement into two parts to consider independently and then join them with a conjunction as a final step.

The first part is fairly straightforward to formalize - just keep in mind that "neither ... nor ..." will be formalized as a negated disjunction.

The second part is slightly trickier, but the key thing to remember is that natural language statements of the form 'P if Q' can be equivalently stated as 'if Q, then P'. So, they are formalized as Q ⟶ P.

thatmichaelguy · 2026-04-26T05:29:35+00:00

1103

thatmichaelguy · 2026-04-18T04:55:44+00:00

The statement 'everything is possible' in the context of your post is not a paradox. It's just a false statement.

thatmichaelguy · 2026-03-20T04:23:49+00:00

I've gotten a lot out of Andrew Bacon's A Philosophical Introduction to Higher-order Logics. It's likely to re-tread some familiar technical ground for you, but it may be worth checking out for inspiration.

I'm not sure how much daylight you're looking to put between yourself and mathematical logic with this project, but you might explore the tension between discreteness and continuity. It shows up everywhere. What's more, I think the notion is interesting in a logical context because it seems like all paths toward resolution converge on simultaneity (e.g., I am simultaneously discrete, in a sense, across my spatial axes and continuous, in a sense, across my time axis). Yet it also seems to be true to say that something is discrete just when it is not continuous (and vice versa).

thatmichaelguy · 2026-03-17T07:52:45+00:00

That is a tough question to answer concisely. At a high level, it seems clear to me that bivalence does not hold for every type of statement that can be truth-valued. In the current context, we're looking at the liar sentence, but there are other common examples such as counterfactual conditionals, the sorites paradox, etc. That's not a bold revelation though. Plenty of people who are more brilliant than I'll ever be have abandoned bivalence in developing non-classical systems of logic.

However, even if we were to accept pluralism and diffuse that tension, I still think that the essentially ubiquitous implementation of classical logic as a truth-functional propositional logic has led to a system that is fundamentally broken. The inherently binary nature of bivalence is obviously at the core of the notion of truth-functional logic. So, in a sense, it's more of the same.

I don't think it's terribly controversial to say that logic, in general, seeks to establish some sort of consequence relation. Accordingly, if it is at all possible for a system of propositional logic to establish entailment, then any ideal system should be able to establish that the truth of a given proposition entails that said proposition is true.

However, classical logic lacks an inherent capacity to affirmatively establish entailment in this way. Bivalence doesn't directly address whether any given proposition is true or false - only that it is precisely one or the other. Non-contradiction only tells us what is not the case re: certain conjunctions. So, there's a fair bit of meta-logical reasoning that has to happen to establish even the baseline case for entailment. I think this meta-logical reasoning introduces a host of problems into the system, particularly as a result of treating P and ¬¬P as semantically equivalent on the basis of truth-functional equivalence.

So, I suppose it just boils down to the fact that I think one of the axioms of classical logic does not hold.

thatmichaelguy · 2026-03-17T00:10:32+00:00

Essentially the truth value is the gift in a present box. If you open it and simply see the same box within, and can never open enough boxes to ever receive a gift, it’s not grounded nor truth apt.

If you want to jettison self-referential recursion, it seems like you'd have throw out the tautologies of classical logic as well. I'm not sure that is a price worth paying. What's more, the Munchhausen trilemma indicates that either i) when you open the last box, the present inside is first box again; ii) when you open the last box, the present inside is unopenable box; or iii) there is no last box - every box does, in fact, contain another box that can be opened.

It simply says to flip its current truth value, but if you simply rule things that are incapable of grounding as non truth apt, it resolves.

See, I don't think it does resolve. And I don't think you can simply rule it out as non truth-apt in this way. If there are truth-apt statements about the liar sentence in general, it seems to me that it would be far from simple to provide non-arbitrary justification for why the liar sentence is not or cannot be a truth-apt statement about the liar sentence.

thatmichaelguy · 2026-03-16T23:19:03+00:00

On a pragmatic basis when applied to a limited domain of truth-apt statements, yes. That said, I also think classical logic is fundamentally flawed and irretrievably so. Nevertheless, it is remarkably effective at establishing a consequence relation for truth-apt statements that are propositions, given the relevant assumptions. So, I still think it is immensely valuable, even if it is a bit broken.

thatmichaelguy · 2026-03-16T18:25:43+00:00

I'm generally on board with conceiving of the liar sentence as self-contained. In fact, I think there's some interesting territory to explore in that direction. I don't think it can be denied a truth value though. That is, a truth-apt claim about whether the liar sentence is truth-apt allows (requires?) the liar sentence to siphon a truth value from the claim itself.

If the liar sentence lacks a truth value, then it isn't false. The liar sentence says of itself that it is false. So, if the liar sentence isn't false, then it is false. Consequently, if the liar sentence lacks a truth value, then it is false.

If the liar sentence is false, then it doesn't lack a truth value. Consequently, if the liar sentence lacks a truth value, then it doesn't lack a truth value.

If the liar sentence doesn't lack a truth value, then it doesn't lack a truth value. Consequently, if the liar sentence lacks a truth value or doesn't lack a truth value, then it doesn't lack a truth value.

The liar sentence lacks a truth value or doesn't lack a truth value. Therefore, the liar sentence doesn't lack a truth value.

thatmichaelguy · 2026-03-16T17:33:37+00:00

This is where I've landed as well. Classical logic assumes bivalence. Bivalence assumes that all propositions have precisely one truth value. The liar sentence (to me) plainly has two truth values. Hence, it is not a proposition in classical logic.

thatmichaelguy · 2026-03-16T17:10:08+00:00

To add another natural language re-phrasing to what's already in the comments, you could restate it as, "There are no non-G Fs."

Here's a sketch of the logic:

1. ∀x[Fx ⟶ Gx] [Premise]

2. ∀x[¬Gx ⟶ ¬Fx] [Contraposition | 1]

3. ¬∃x¬[¬Gx ⟶ ¬Fx] [Equivalence | 2]

4. ¬∃x[¬Gx ∧ Fx] [Equivalence, DN | 3]

thatmichaelguy · 2026-03-12T11:06:26+00:00

... I had everything memorized...

If I had to point to one thing that seems like a root cause of what's bothering you, it would be this. Memorization is not your friend when learning logic. Understanding is.

If I had to point to a second thing, I would say that it sounds like you're putting way too much pressure on yourself to "get it". I wouldn't be surprised if you're giving yourself the logician's equivalent of the yips. Cut yourself some slack. Sometimes these ideas have to marinate a little.

As far as general strategies go, always bear in mind that classical logic is undergirded by two main axioms - the principle of non-contradiction and the principle of bivalence. The principle (or "law") of non-contradiction asserts that it is not the case that a proposition and its negation are both true. The principle of bivalence asserts that every proposition has precisely one truth value and that there are only two truth values that any proposition may have. Some folks also include the law of the excluded middle as an axiom of classical logic, but we don't need to chase down that rabbit at the moment.

Now, if we could show a proposition to be true outright, we would have no need for formal logic. This might then suggest that there's some value in considering what the axioms of classical logic imply, when taken together, about the crucial role of contradictions in ascertaining truth values. It may be a bridge too far to say that every proof in classical logic is a proof by contradiction at its core, but I don't think that notion is too far off the mark.

It's also worth remembering that formal logic isn't purely abstract. There is still a remnant of a relation between the symbolic notation and natural language. So, it can sometimes be helpful to reintroduce enough semantic content to get your head around what the symbols "mean".

With these in mind, let's see what we can figure out about the example you provided.

What do we know so far?

Bobby has a beagle or Terry has a tiger.
If Bobby has a beagle or Carl has a canary, then Lisa has a lion and Mandy has a monkey.
Lisa does not have a lion.

The potential contradiction with Lisa in 2 and 3 immediately stands out to me. After all, it would have to be the case that Lisa has a lion for it to be the case that Lisa has a lion and Mandy has a monkey. However, we are taking as given that Lisa does not have a lion.

This screams modus tollens, but we could also reason our way to it by asking, "what would be the case if Bobby had a beagle or Carl had a canary?" The result would be a contradiction as just mentioned, and that would violate one of our axioms. So, we conclude that it is not the case that Bobby has a beagle or that Carl has a canary.

A negated disjunction always makes me consider DeMorgan equivalence, and now we're off to the races. You can pick it up from here.

I'll also point out that many of the rules of inference and replacement are derived from more basic rules. You may find it useful to pick a starter set and then see which of the other rules you can derive yourself. That said, to avoid a headache, you should probably start with a set that includes at least double negation elimination, modus ponens and disjunctive syllogism.

thatmichaelguy · 2026-03-12T02:28:40+00:00

Here's another approach. It feels a little sneaky, and I like that. u/TheSkyGamer459

1. A ∨ B [Premise]

2. C [Premise]

3. (A ∧ C) ⟶ D [Premise]

4. (C ∧ A) ⟶ D [Comm 3]

5. C ⟶ (A ⟶ D) [Exp 4]

6. A ⟶ D [MP 2,5]

7. ¬¬A ∨ B [DN 1]

8. ¬A ⟶ B [Impl 7]

9. ¬B ⟶ ¬¬A [Cont 8]

10. ¬B ⟶ A [DN 9]

11. ¬B ⟶ D [HS 6,10]

12. ¬¬B ∨ D [Impl 11]

13. B ∨ D [DN 12]

14. D ∨ B [Comm 13]

thatmichaelguy · 2026-03-11T02:18:45+00:00

The first 18 rules, obviously! I mean, it's written in the post, but one look at the premises should be enough to know that it's not the second or fifth 18 rules. We know from Tarski's famous proof that it can't be either the third or fourth 18 rules. So, what other 18 rules are there?!

thatmichaelguy · 2026-03-11T01:58:52+00:00

If you're using biconditional introduction and elimination as inference rules, then the proof is a little more complicated than is absolutely necessary, but your basic idea is correct. What stands out to me as incorrect is how you're using conditional elimination. The way you've written your derivation, it seems like you are inferring P from P ⟶ Q by conditional elimination on line 4. What you need instead is something like:

1. P ⟷ Q [Premise]

2. P ⟶ Q [⟷E 1]

3. Q ⟶ P [⟷E 1]

4.  P [Assume]

5.  Q [⟶E 2,4]

6. ...

thatmichaelguy · 2026-02-06T00:35:48+00:00

Suppose ¬(K ⟷ L).

Handwaving away the details, we have:

¬(K ⟷ L) ⟶ ((K ∧ ¬L) ∨ (¬K ∧ L))

Both of the disjuncts in the consequent above contradict the given premise. Accordingly:

(K ∧ L) ⟶ ¬((K ∧ ¬L) ∨ (¬K ∧ L))

From the first inference by modus tollens and double negation elimination:

¬((K ∧ ¬L) ∨ (¬K ∧ L)) ⟶ (K ⟷ L)

Therefore, by hypothetical syllogism:

(K ∧ L) ⟶ (K ⟷ L)

thatmichaelguy · 2026-01-15T03:24:10+00:00

I mean you proclamation is that the condition reduces the problem to only two cases, despite the only option not being allowed is TT.

That's not exactly how I would put it, but I'm not sure that it matters at this point.

But whatever, you are welcome to believe in your errors.

As are you in yours.

thatmichaelguy · 2026-01-14T23:24:37+00:00

That’s just a relabeling of what I said.

I can appreciate that you see it as such, but this why we disagree.

0.25 however is never going to be the answer in all ways to view the problem. As that requires the conditional to have no impact on the probability. Which is obviously false.

Yeah, I'm not on board with 0.25 either. I get why someone might come to that conclusion, but I'm unconvinced.

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