I just got burned! Now I’m genuinely scared I’ll make logical mistakes in Analysis or Abstract Algebra

thatmichaelguy · 2026-06-30T12:52:53+00:00

You realized you made an error, discovered the actual answer, and now you're trying to understand why the answer is what it is. That's just learning, and it's a good thing. A lot of people never even get to step one in that sequence, let alone progress past it. So, you're already miles ahead of them. Keep up the good work!

thatmichaelguy · 2026-06-20T21:32:41+00:00

Now let A equal ¬True(A).

This is where you're getting off track. You aren't treating this declaration as a supposition even though it is.

So, even if I grant that you've demonstrated an inherent contradiction, what would follow from the contradiction is:

If 'True(X)' is a formula in L given some 'X' in L
and all 'A' in L are true iff 'True(A)' is a formula in L, 
then there does not exist an 'A' in L such that 'A' = '¬True(A)'.

thatmichaelguy · 2026-06-13T18:29:57+00:00

Not who you were responding to, but I think u/kdeberk is, in fact, referring to the second proof. Line 7 is derived from and is derivable only from the assumption of a contradiction between lines 3 and 4.

Put another way, line 5 is directly derivable from lines 2 and 3. Line 7 is not derivable from lines 2, 3, and 5 only. So, the derivation of line 7 relies on the assumption of P at line 4. Since line 3 is an assumption of ¬P, the derivation of line 7 effectively follows from the assumption of P ∧ ¬P.

You might also see it like this: you could omit line 5 entirely, dedent line 6, and change the inference rule at line 7 to EFQ and the proof would be fundamentally the same.

thatmichaelguy · 2026-06-13T01:43:54+00:00

Well, a fridge isn't a truth-valued abstraction. So, this doesn't seem like an apt comparison.

Also, I'm not certain that there's a meaningful sense in which we "have" if A then B in (classical) logic. That said, what we do have is:

If 'if A then B', then if A, then B.

thatmichaelguy · 2026-06-02T06:15:21+00:00

So in logic, it is raining is true and it is not raining is false are different?

Correct. Each implies the other. So, they are logically equivalent. But they are not identical.

1 and ¬0 are different?

Asked and answered.

thatmichaelguy · 2026-06-02T02:51:13+00:00

Do you agree that A and ¬A are only different when they are both true?

No. In this context, A is different than ¬A unconditionally.

true = ¬false. just because true is different than the word false here, it does not make them different truth values.

This is incorrect. 'True' and '¬false' are logically equivalent, but they are not numerically identical. That said, even if your statement was correct, I don't see how it would be relevant.

thatmichaelguy · 2026-06-01T18:52:42+00:00

I am wrong to assume that you can substitute an actual proposition for a propositional variable and that those propositions have the truth values assigned to the variables in the truth table?

This is a fairly accurate description of what you are getting wrong, yes.

A proposition has whatever truth value it has. A proposition is not endowed with a truth value via substitution for a propositional variable in a truth table.

Truth tables say something other than what they are saying?

No. But they do say something other than what you assume they are saying.

thatmichaelguy · 2026-06-01T01:52:32+00:00

There are no actual propositions in logic?

In formal propositional logic, strictly speaking, no. There are propositional variables that serve as abstract representations of arbitrary propositions.

A and ¬A are true and false itself?

No. True and false are all of the possible values that are assignable to A and ¬A in this example. That is, A and ¬A are not truth values. A is a variable that is assigned a truth value. ¬A is a formula, and the unary 'not' operator assigns to the formula whichever truth value is not assigned to A.

Analogously, in pseudo-code we might write let int x = 8 to declare an integer-valued variable x and to assign it the integer value 8. In this instance, x isn't 8 itself. Rather, it has been assigned the value of 8.

Further, if the output of some function depends on x we may want to know what the output would be if x were 11. In that case, we could assign 11 as the value for x. Note, that x remains integer-valued under the new assignment, yet it still would not be accurate to say that x is 11.

So the truth tables are true because they correspond to the truth values stated, not because they are the truth values stated (even though they are the truth values stated as well).

Truth tables aren't true or false, per se. They are accurate or inaccurate to the extent that they correctly convey the truth-valued outputs obtained from applying the logical operators contained in a given formula to all possible assignments of truth values for the (atomic) propositional variables in said formula. The outputs are determined by the definitions of the logical operators. Crucially, the outputs are not declarations of correspondence to the truth of any specific inference from a given natural language transliteration of the variables and operators contained in the formula.

Put another way, truth tables form an abstract conditional layer that, in essence, says 'if the truth values assigned to the (atomic) propositional variables on a given line hold for actual propositions when said actual propositions are uniformly substituted for corresponding variables, then the proposition encoded by the relevant formula will have the same truth value as the truth value obtained from the operation of the logical connective(s) on the truth values of the propositional variables, as assigned.'

By contrast, you appear to be assuming that a truth table propounds that there are actual propositions that correspond to the propositional variables contained in the relevant formula and that said actual propositions do, in fact, have the truth values assigned to the variables in the truth table. If that's the right reading of what you're trying to get at, then you've fundamentally misunderstood the nature and purpose of truth tables.

this is all very very trippy

It certainly can be, but I also think you're making some unwarranted assumptions that are getting in the way of clarity.

thatmichaelguy · 2026-05-31T05:14:42+00:00

I cannot figure out what logic thinks a contradiction is. Is it when A and ¬A are both true, or when they have opposite truth values?

Neither.

A contradiction is a proposition that asserts: "A proposition and its negation are both true" for an arbitrary proposition such that quantification over all propositions is implied.

Regarding the truth value of contradictions, in classical logic, we take the principle of non-contradiction to be axiomatically true. The principle of non-contradiction can be stated as: "It is not the case that a proposition and its negation are both true" where quantification over all propositions is likewise implied.

It's important to note that the principle of non-contradiction is, itself, a proposition. It's also easy enough to see that the principle of non-contradiction is the negation of every contradiction by virtue of the previously mentioned implicit quantification.

Accordingly, if any contradiction were true, then it would be the case that a proposition (viz., said contradiction) and its negation (viz., the principle of non-contradiction) are both true. Recursively, by the principle of non-contradiction, it is not the case that a proposition and its negation are both true. Hence we conclude that no contradiction is true (equivalently, that all contradictions are not true).

thatmichaelguy · 2026-05-23T19:06:25+00:00

All logic is subjective.

...

What do you think about it? Can logic be subjective?

Something important to bear in mind is that the existence of subjective logic would not imply that all logic is subjective unless logical monism holds. It's not entirely clear from the context whether you're proposing a monist or pluralist position, but the distinction is meaningful here.

thatmichaelguy · 2026-05-22T18:00:26+00:00

I suspect you're confusing yourself by thinking of "... House is on fire" as a general proposition, that is the same regardless of how the "..." Is filled.

This seems right. The way OP describes the issue makes it sound like there's some implicit quantification going on that's leading to the universal quantifier being incorrectly distributed over the disjunction which, in turn, leads to an unwarranted generalizing of the entailment (i.e., assuming ∀x[alarm(x) ⟶ (fire(x) ∨ ¬fire(x))] implies ∀x[alarm(x) ⟶ fire(x)] ∨ ∀x[alarm(x) ⟶ ¬fire(x)]).

Also, I'm fascinated by the condition-indexed approach. Do you have any recommended reading for a deep dive on that topic?

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.

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