Cheers. The wider point is that formal logic removes ambiguities in natural languages, which is what I take you have been saying.

1. Right, I mean P entails Q. The logical equivalence (not-P or Q) is an implication of the conditional, not having the same meaning as the conditional. — NotAristotle

Where G is a set of sentences and Q is a sentence, "G entails Q" is symbolized:

G |= Q

I.e, there is no interpretation in which all the members of G are true and P is false.

If G is a singleton {P}, then we sometimes write:

P |= Q

I.e., there is no interpretation in which P is true and Q is false.

Then note:

P -> Q |= ~P v Q
and
~P v Q |= P -> Q [corrected in edit]

Whatever you mean by "meaning", the sentences are equivalent in the sense above.

2. I take your question to be what would a rule be, how is it defined? I would define a rule as a member belonging to a set that exhausts all "truth possibilities." I would add that the following of a rule may not result in a contradiction. — NotAristotle

See later in this post for my reply to your supposed rules given lately.

Member of what sets? What sets are you talking about?

And what does "exhausts all truth possibilities" mean?

And an application of rule may not result in a contradiction? You said previously that you don't define 'correct' for rules. So consider the rule of conjunction-intro:

From P and Q infer P&Q.

If P is A and Q is ~A, then apply the rule to get A & ~A.

Conjunction-intro is not a rule for you now?

A rule relating two different variables would have (I think) 15 possible truth configurations. — NotAristotle

There are 16 2-place Boolean functions.

The rules must at least enable all those possibilities to be instantiated (though perhaps it may exclude possibilities that are necessarily contradictory). — NotAristotle

What does it mean to "instantiate" in that regard?

What we do have is this:

All 16 2-place Boolean functions are realized by certain sets of connectives. And if all 2-place Boolean functions are realized, then all n-place Boolean functions are realized for all n.

That is not regarding inference rules, but merely the definitions of the connectives and the truth evaluation of compound sentences.

3. "Some proposition is not the case"
Both propositions must be true
Either proposition must be true
If the one proposition is true, so must the consequent proposition
Both propositions are either both true or both false. — NotAristotle

Those are not rules for arguments. Those are just the standard clauses in the ordinary definition for the truth value of compound sentences. (Except your use of "must", and note that you left out "must" for the first and fifth.)

And it's not clear what you mean. Do you mean "must be true" as "is necessarily true"? With modal operator n for 'necessarily', taken at face value, your formulations seem to be:

~P is true if and only if P is false. (So far, so good!)

P & Q is true if and only if both nP and nQ.

P v Q is true if and only if either nP or nQ. (And, unless you tell me otherwise, I take 'or' as inclusive).

P -> Q is true if and only if P necessarily implies Q. (?)

P <-> Q if and only if either both P and Q are true or both P and Q are false. (Good!)

5. Valid argument = following the rules, where rules are defined as those operations that enable each truth possibility to be instantiated but that do not result in a contradiction by following that rule. — NotAristotle

What does "enable truth possibilities to be instantiated" mean?

And what if a rule allows a contradictory sentence (but that itself is not a contradiction) to be derived? Note that in predicate logic there is no mechanical procedure to determine whether any given sentence is or is not contradictory. So, with your offer, there would not be a mechanical procedure to determine whether an argument did use only your rules. But, of course, we may consider such a logic.

8. Not logical anarchy; the rules must enable all truth possibilities to be instantiated except that the rule may not result in a contradiction if it is followed. — NotAristotle

That might be helpful if you define "enable all truth possibilities to be instantiated" vis-a-vis rules.

This way of defining validity may be preferable because it deals with cases such as A->not-A therefore Not-A that are intuitively illogical; such an argument does not involve the following of a rule, and so it is not valid. — NotAristotle

It doesn't derive a contradiction. So in what way does it fail to "enable all truth possibilities to be instantiated"?

Similarly, A, A->not-A therefore not-A another intuitively illogical seeming argument would not be valid because the following of the rule results in a contradiction. — NotAristotle

No, it doesn't result in a contradiction. The conclusion is ~A, which is not a contradiction. Yes, the premises are inconsistent, but your definition of "rule" doesn't disallow inconsistent sets of premises, only required is that application of the rule doesn't allow a conclusion that is a contradiction. The particular application you mentioned doesn't derive a contradiction. But other applications do derive contradictions. I know of no rule that disallows deriving contradictions, since rules don't disallow inconsistent sets of premises. However, as mentioned previously in this thread, one might require rules to not have inconsistent sets of premises. But the catch ... for predicate logic, it would not be algorithmically checkable to see whether a rule was applied, since it is not algorithmically checkable to see whether a set of sentences is inconsistent. Though, I suppose you mean for all your rules to be informal anyway.

...but it does not have the same meaning qua modus ponens (or modus tollens) — Leontiskos

What does that mean?

https://www.umsu.de/trees/#(((A~5B)~1A)~5B)~4(((~3A~2B)~1A)~5B)

Wow. What a post.

A -> B
A
therefore, B

is not the same argument as

~A v B
A
therefore, B

That doesn't vitiate that A -> B and ~A v B are equivalent.

"P does not imply Q".

Depends on what 'implies' means.

It is not the case that if P then Q
is formalized
~(P -> Q)
that's in the object language

It is not the case that P entails Q
is formalized
P |/= Q
that's in the meta-language

the full meaning of P->Q — NotAristotle

According to you, what is the full meaning of P -> Q?

Right.

But is there anything more to it than the difference in the shapes of the letters?

OK, yes, same question. Nevermind to my above.

the "meaning" of the disjunctive is not specific enough — NotAristotle

The ordinary clause is:

P v Q is true if and only if either P is true or Q is true. ('or' inclusive)

P -> Q is true if and only if either P is false or Q is true ('or' inclusive)

What specificity is lacking?

I was disputed that the following definitions are equivalent versions of the ordinary textbook definition:

(1) An argument is valid if and only if there are no interpretations in which all of the premises are true and the conclusion is false

(2) An argument is valid if and only if every interpretation in which all of the premises are true is an interpretation in which the conclusion is true

The dispute comes down to a claim that, without justification I had applied the material conditional in the meta-language.

I have explained that in ordinary formal logic, the material conditional is used for both the object language and meta-language, especially since the meta-language itself is formalizable and logicians don't eschew the material conditional merely on account of moving to the meta-language.

I cited numerous definitions of 'valid', some using (1) and some using (2). They are equivalent, though a writer may choose one or the other and not mention the other or the equivalence, since the material conditional is indeed the sense of "if then" in ordinary formal logic, whether object-language or meta-language.

But I happened to come across a text that does mention the equivalence:

"A set of sentences G implies or has a consequence the sentence D if and only if there is no interpretation that makes every sentence in G true, but makes D false. This is the same as saying that every interpretation that makes every sentence in G true makes D true." - Computability And Logic - Boolos, Burgess and Jeffrey.

(Very minor and not material technical disclaimer: The quote above is followed by a technical exception, but also an explanation of how that technical exception dissolves upon understanding that 'every interpretation' may be taken to mean 'every interpretation that assigns denotations to all the nonlogical symbols in whatever sentences we are considering'.)

Whether or not the two expressions are semantically equivalent in a meta-logical sense depends on how one is using them. — Leontiskos

Hmm interesting, I think my position is that the formal conditional is meaningless then, insofar as it is just symbol manipulation.

You could say that, but you would end up having to admit that "P does not imply Q" cannot be formalized in any way whatsoever, at least in propositional logic. — Leontiskos

I have tried to formalize it and can't seem to do so; this is an approximation:

(A v ~A) → (~B v ~A)

When (B and A) are both true, the expression seems to be false. On the other hand, the negation of that expression seems to imply that (A and B) must both be true. If the conditional is construed as only being true when A and B are true, then the negation of the initial expression maps onto A→B. Perhaps that could be written as, it is not the case that A does not imply B therefore A implies B. (Though if that were the case then A→B would be logically equivalent to A^B, although not meta-logically equivalent).

But then I don't mind saying "P does not Imply Q" can't be formalized.

According to you, what is the full meaning of P -> Q? — TonesInDeepFreeze

I may have mispoken, but to me the full meaning of "If P then Q" captures the fact that "P does not imply Q" can still be true even though not-P v Q can still be true. But then I now think P->Q is a meaningless expression so saying it "means" the same think as not-P or Q is unsubstantiated.

What are the "full meanings" of "If P then Q" and "P does not imply Q", according to you?

And what is the difference, according to you, between "the meaning" and "the full meaning"?

By "the following of a rule" I mean a literal rule such as a connective is actually used to reach a conclusion. The argument A->not-A therefore not-A does not, in my opinion, make any use of the conditional such that any rule has been followed. With the argument A->not-A, A, therefore not-A, the following of the rule, namely the conditional in that argument, leads to a contradiction between A and not-A, as such, it is disqualified from being a valid argument according to my definition.

You still have not stated any rules.

The argument:

A -> ~A
therefore A

makes use of the interpretative clauses for '->' and '~'.

But I have not mentioned a rule, since the above is merely an argument and not a proof.

With the argument A->not-A, A, therefore not-A, the following of the rule, namely the conditional in that argument, leads to a contradiction between A and not-A — NotAristotle

You didn't read a word I wrote about that.

I have tried to formalize it and can't seem to do so; this is an approximation:

(A v ~A) → (~B v ~A) — NotAristotle

That's very incorrect.

Why don't you just read one chapter in an intro textbook? Is there some reason you won't read even a few pages of a book or article to inform yourself on the subject? Are you allergic to books and articles or something? Have a phobia of them or something?

No, I read it, I just think you're disregarding the proviso I stated, namely that a rule must actually have been followed, not merely be present in an argument.

As for the instantiation of truth possibilities by the rules, what I mean is that the possibilities for what is true and what is false are arrayed across a truth table. The rules must account for all the ways that those truth possibilities can be instantiated. So for the expression A v B, the truth table is T, T, T, F. On the other hand, T, F, F, F, is A ^ B. Every possibility wherein T is present must be uniquely accounted for by the rules. So T, F, F, F, and F, T, F, F, and F, F, T, F, and F, F, F, T, must all be "achievable instantiations" based on the rules we bring to the variables. If A v B were the only rule we applied, then not all of the truth possibilities could be instantiated, does that answer?

If the conditional is construed as only being true when A and B are true — NotAristotle

Is that your offering? So, the conditional is false when A is true and B is false (that is ordinary), when A is false and B is true, and when A is false and B is false?

Then we would have A -> B is true if and only if A & B is true, So "if then" to you is just conjunction?

Not just conjunction, no, but having the same truth functionality as conjunction yes, just meta-logically different (if I am using that terminology correctly).

Then note:

P -> Q |= ~P v Q
and
~P v Q |= ~P v Q — TonesInDeepFreeze

I think you meant:

P -> Q |= ~P v Q
and
~P v Q |= P -> Q

?

I just think you're disregarding the proviso I stated, namely that a rule must actually have been followed, not merely be present in an argument. — NotAristotle

What is the difference between "followed" and "present"? And, according to you, how does it vitiate what I said about contradiction.

what I mean is that the possibilities for what is true and what is false are arrayed across a truth table. — NotAristotle

The ordinary rules do not eschew any rows in any truth table.

So for the expression A v B, the truth table is T, T, T, F. On the other hand, T, F, F, F, is A ^ B. Every possibility wherein T is present must be uniquely accounted for by the rules. — NotAristotle

You didn't read a word I wrote about realizability.

A v B were the only rule we applied — NotAristotle

A v B
is a sentence, not a rule.

Yes, my typo. Thank you.

You're slipping Tones.

Down the slippery slope of formalized illogicality.

You're slipping Tones. — NotAristotle

While you are slipshod.

Down the slippery slope of formalized illogicality. — NotAristotle

You've not shown any illogic I've committed. Meanwhile, you are slippery mess of informal illogic.

just meta-logically different — NotAristotle

What, according to you, is the difference? Yes, they are different formulas, but equivalent according to your own definition. Just waving "meta-logic" like some kind of magic wand is nothing.

The ratio of my substantive typos to good information and good arguments is 1:10000.

The ratio of your confusions, circularity, and ignorance to good information and good arguments is 1:1.

If I am referring to the right quotation, you said:

No, it doesn't result in a contradiction. The conclusion is ~A, which is not a contradiction. Yes, the premises are inconsistent, but your definition of "rule" doesn't disallow inconsistent sets of premises, only required is that application of the rule doesn't allow a conclusion that is a contradiction. The particular application you mentioned doesn't derive a contradiction. — TonesInDeepFreeze

What I responded with --a rule must have been "followed" not merely be "present" and the use of a rule may not result in a contradiction means that the use of a rule, or I guess you would call it an operator or connective, whatever you call it, must not result in a contradiction. A->not-A, when this rule is applied and followed, that is, when it is true that "A" and the rule "A->not-A" is actually applied, a contradiction results, specifically "A and not-A."

By "actually applied" I mean that the rule, or connective, does work in leading to the conclusion.

The "following" of a rule versus it's being merely "present" can be illustrated by the following example:
A->B
B^C
Therefore, C.
In this example, the rule A-> B does not do any work, so even if it did result in a contradiction, the fact that it doesn't do any work in the argument and isn't followed or actually applied, means that the argument could still be valid.