First, why do you ask?

The statement you seem to have in mind is:

(1) It is not the case that there is an interpretation in which all the premises are true and the conclusion is false.

One way to write that:

Let Px stand for "x is an interpretation in which all the premises are true".

Let Qx stand for "x is an interpretation in which the conclusion is false".

Then the statement is:

(2) ~Ex(Px & Qx)

But neither (1) nor (2) capture the definition of 'is a valid argument', which, if we spell out the quantifiers is:

For all T(T is a valid argument
if and only if
(T is an argument
and
for all N(if N is an interpretation, then it is not the case that
((for all p(if p is a premise of T, then p is true per N))
and
for all c(if c is the conclusion of T then c is false per N)))))

Symbolized:

Let Vx stand for x is a valid argument (the definiendum)
Let Bx stand for x is an argument
Let Dx stand for x is an interpretation
Let Rxy stand for x is a premise of y
Let Txy stand for x is true per y
Let Uxy stand for x is the conclusion of y
Let Fxy stand for x is false per y

AT(VT
<->
(BT
&
AN(DN ->
~((Ap(RpT -> TpN))
&
Ac(UcT -> FcN)))))

(I hope I got that all correctly, including the parentheses.)

Why? — Leontiskos

Because of the reasoning explained here.

- Which is a presentation of the principle of explosion, is it not?

Yes

And so you are appealing to or presupposing the principle of explosion when you claim that Susie's argument is valid, are you not?

See the “⊢ Q” at the end? That means that Q follows from the bit before.

We’ve already established that “(P ∨ Q) ∧ ¬P” is true, so therefore “Q” is true.

I don’t know what you mean by “presupposing” the principle of explosion.

I don’t know what you mean by “presupposing” the principle of explosion. — Michael

Do you know if you are appealing to the principle of explosion? Because I asked if you are "appealing to or presupposing the principle of explosion."

I don’t know what you mean by “appealing” to the principle of explosion.

It’s like saying that we “appeal” to modus ponens.

We use modus ponens to derive some conclusion and we use the principle of explosion to derive some conclusion.

This is all a priori reasoning based on logical axioms, not some a posteriori proposition that is possibly false.

- Of course we appeal to modus ponens.

I asked why you think Susie's argument is valid. You gave an argument which you admitted is a presentation of the principle of explosion. Clearly you think Susie's argument is valid because of the principle of explosion, just as we might think that a conclusion follows because of modus ponens. Someone who makes an inference based on a rule of inference is appealing to that rule of inference. This should not be so hard.

An inference is presupposed when it is interpreted as the tacit reasoning of an enthymeme. You are supplying Susie's argument with a rule of inference that she does not explicitly present. You are interpreting it as an enthymeme and supplying what you see as the implicit inferential steps.

"See the “⊢ Q” at the end? That means that Q follows from the bit before." Okay; can you spell it out for me? It's still not clicking.

Here's how I would write it:

1. P & ~P (premise)
2. P (from 1, conjunction elimination)
3. P v Q (from 2, disjunction introduction)
4. ~P (from 1, conjunction elimination)
5. ((P v Q) & ~P) -> Q (theorem)
6. ((P v Q) & ~P (from 3, 4, conjunction introduction)
7. Q (from 5, 6, modus ponens)

Or, have explosion as either a primitive rule or derived rule in a natural deduction system:

1. P & ~P (premise) {1}
2. Q (explosion) {1}

A or B
Not-A
Therefore, B (disjunctive syllogism)

Okay, Thanks for writing out that definition using quantifiers. So could I simplify your argument by saying

E↔A∧(B→¬(C∧D)) is the definition. I know that if we are being precise it is not, but thematically would this work for the definition of validity.



But that doesn't work if A and not-A are both true. That's my point. The proof doesn't work. The proof only works if you ignore that A is also true.

I can only guess, but I think that is what Tones meant by referring to that step as a "theorem."

E↔A∧(B→¬(C∧D)) — NotAristotle

That is not a rendering of my formulation.

Alright, how might you render it in a simplified form, or can it not be so rendered?

I adduced it as a previously proven theorem to help you see how the final step would by an application of modus ponens, so you'd have another way to look at it to see that the steps are correct. We could also do it this way:

Rule (disjunctive syllogism): If P v Q occurs on a line, and ~P occurs on a line, then infer Q.

1. P & ~P (premise)
2. P (from 1, conjunction elimination)
3. P v Q (from 2, disjunction introduction)
4. ~P (from 1, conjunction elimination)
Q (from 3, 4, disjunctive syllogism)

The definition involves quantification, so I wouldn't reduce it to a merely sentential formula.

Yeah, I don't get how you get Q from (P or Q) if P is true. And I understand the disjunctive syllogism. I get that your asserting not-P, but I don't see how that negates a proposition, P, that has been stipulated to be true, per the argument.

But that doesn't work if A and not-A are both true. — NotAristotle

(1) In no interpretation are both A and ~A true.

(2) Having A & ~A as a premise, thus being able to have A as a line and ~A as a line, does not vitiate use of disjunctive syllogism:

The rule is: If P v Q is on a line, and ~P is on a line, then infer Q.

The rule is NOT: If P v Q is on a line, and ~P is on a line, and P is not on a line, then infer Q.

But that doesn't work if A and not-A are both true. That's my point. The proof doesn't work. The proof only works if you ignore that A is also true. — NotAristotle

Sure, and that's the same puzzle of the OP. I see your point.

See:

In cases of inconsistent premises what happens is that the person arguing arbitrarily makes use of some premises while conveniently ignoring others. For example:... — Leontiskos

I don't get how you get Q from (P or Q) if P is true. — NotAristotle

In this case:

We have the premise P & ~P.

We want to get P v Q and we want to get ~P, so we can apply disjunctive syllogism to get Q.

We get P v Q by first getting P from P & ~P by conjunction elimination, then P v Q from P by disjunction introduction.

We get ~P from P & ~P by conjunction elimination.

/

The rule is: If P v Q is on a line, and ~P is on a line, then infer Q.

The rule is NOT: If P v Q is on a line, and ~P is on a line, and P is not on a line, then infer Q.

The rule is: If P v Q is on a line, and ~P is on a line, then infer Q.

The rule is NOT: If P v Q is on a line, and ~P is on a line, and P is not on a line, then infer Q. — TonesInDeepFreeze

I think @NotAristotle is right insofar as the rule is ambiguous. There is no magical rule-book of logic that settles this issue, and in practice someone who contradicts themselves is responded to with a reductio.

This should not be so hard. — Leontiskos

You’re right, it shouldn’t. Which is why I don’t understand why you are taking issue with what I am saying.

It is simply an a priori fact that from “p and not p” one can derive any conclusion, and so any argument with “p” and “not p” as premises is valid.

The rule is completely unambiguous:

If P v Q is on a line, and ~P is on a line, then we may put Q on a new line.

Or better, without "we", "may" and "put", the rule may be stated:

A deduction from a set of formulas G is a sequence of formulas such that:

If a formula P appears on a line, then it is either a member of G or there are formulas on previous lines such that there is a rule such P comes from those formulas.

And among the rules is: P v Q, ~P |- Q.

What disjunctive syllogism is is settled by any ordinary textbook that has it as a rule. Whatever particular wording is used in an ordinary textbook, it amounts to the rule that from P v Q and ~P we may infer Q.

I think that is right, it is arbitrary. Although I would say that an argument can have inconsistent premises and still be valid as long as those premises do not do any "work" in the argument, but I acknowledge that my definition of validity may be atypical. At least, I would guess that Tones regards it as unconventional.

It is simply an a priori fact that from “p and not p” one can derive any conclusion, and so any argument with “p” and “not p” as premises is valid. — Michael

You think it's valid because of explosion. It's that simple. Again:

TonesInDeepFreeze thinks that any argument with inconsistent premises is valid, and that the principle of explosion does not need to be presupposed in order to say this. Michael disagrees. He thinks that any argument with inconsistent premises is valid, and that the principle of explosion does need to be presupposed in order to say this. — Leontiskos

• Michael: Susie's argument is valid because of explosion.
• Tones: Susie's argument is valid, but not because of explosion.

What textbook says that. If you can cite that statement I'll sell the farm.

I think that is right, it is arbitrary. Although I would say that an argument can have inconsistent premises and still be valid as long as those premises do not do any "work" in the argument — NotAristotle

Sure, but a premise that is not doing any work in an argument is not a premise of the argument. It is an unrelated proposition. You want to say that if one half a contradictory pair is doing work, then the other half is implicated.

What textbook says that. If you can cite that statement I'll sell the farm. — NotAristotle

I'm sure you could find that in a textbook, but one must recognize that such textbooks presuppose that the premises are not inconsistent.