the more purely formal a system is, the less this discontinuity of reductio ad absurdum is able to be recognized. — Leontiskos
Laws of deduction are not usually derived from one another. — Banno
"A does not imply a contradiction" is not a true statement about "(A→¬(B and ¬B))". — TonesInDeepFreeze
What you say here is blatantly erroneous.First, this is not a derivation of RAA. It is a putative modus tollens that looks a little bit like an RAA. As I said, there are analogical similarities. — Leontiskos
Such derivations have been presented here by several folk, including the one from the IEP given above...there is a measure of discontinuity between RAA and the other inferences of classical propositional logic, such that there is no straightforward derivation of RAA from these other rules of inference.* — Leontiskos
And then repeated them, despite their being shown to be wrong.You say that I have made a number of well-documented errors in this thread. — Leontiskos
...for someone who has failed to engage correctly with formal logic this may indeed be so.* What I have said more recently is that the more purely formal a system is, the less this discontinuity of reductio ad absurdum is able to be recognized. — Leontiskos
In axiomatic systems RAA is derived.So there will be unique and irreducible inference-axioms in any inferential system, but my claim is that RAA is uniquely unique. — Leontiskos
...in a dialogical context (which is my primary context) a MP cannot be rebuffed, but a reductio can.
— Leontiskos
I'll invite you to set out an example. It might be helpful. — Banno
There is nothing "putative" about the use of MP — Banno
Modus Tollens tells us that "Given ψ→ω, together with ~ω, we can infer ~ψ". In the first example you do not have ~ω. It might as well be a Reductio, although even there it is incomplete. — Banno
What is at stake is meaning, not notation. To draw the modus tollens without ¬(B∧¬B) requires us to mean FALSE. You say that you are not using a modus tollens in the first argument. Fair enough: then you don't necessarily mean FALSE. — Leontiskos
Such derivations have been presented here by several folk, including the one from the IEP given above. — Banno
Look at proof 10.5. It is a proof of (~A→A)⊢A in an axiomatic system. — Banno
My bad, I shouldn't have uncritically adopted your nomenclature. Laws of deduction are not usually derived from one another. But deriving equivalent schema to MT and RAA are exercises in basic logic. Here's one using MT:
ρ→(φ^~φ) (premise)
~(φ^~φ) (law of non contradiction)
:. ~ρ (modus tollens)
— flannel jesus
And the conclusion is ρ→(φ^~φ)⊢~p, one of the variants of RAA. — Banno
This is perhaps my favorite proof for the modus tollens thus far. The question is whether that second step justifies the modus tollens. Does the "law of non contradiction" in step two allow us to think of the contradiction as a simple kind of falsity, which requires no truth-assignment? And if so, does that thing (whatever it is), allow us to draw the modus tollens? These are the questions I have been asking for 12 pages.
See my posts <here> and <here> for some of the curious differences between (φ^~φ) and ¬(φ^~φ). — Leontiskos
Here's the thing: nearly every one of your posts in this thread contains factual errors. — Banno
You say that I have made a number of well-documented errors in this thread. This is assertion and hot air which can in no way be substantiated, but there is a way for you to show that my corollary is mistaken. . . — Leontiskos
(NB: Given the way that common speech differs from material implication, in common speech the two speakers would generally be contradicting one another.) — Leontiskos
I understand that you'd think that B∧¬B should be able to be replaced by any proposition P, but that is not the case.
Example:
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid. — Lionino
((p→q)∧(p→¬q)) and (p→(q∧¬q)) are the same formula — Lionino
I understand that you'd think that B∧¬B should be able to be replaced by any proposition P, but that is not the case.
Example:
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid. — Lionino
[Using (b∧¬b) within formulas] is a bit like putting ethanol fuel in your gasoline engine and hoping that it still runs. — Leontiskos
I don't think that's quite right, depending on what you meant by "generally". — Srap Tasmaner
That proposes another link, and I would suggest that in everyday reasoning the truth of (3) requires the falsity of (1), even though P→~Q does not entail ~(P→Q), which indeed does seem to be a problem for material implication. — Srap Tasmaner
I think people do recognize the difference even in everyday reasoning, and would accept that (2) is the simple contradiction of (1), and that (3), while also denying (1) a fortiori, is a much stronger claim. — Srap Tasmaner
If no Englishmen are honorable, then it stands to reason that not all of them are, but that's a much stronger claim than simply denying that being English entails being honorable. — Srap Tasmaner
A conditional, by its very name, signifies that which is not necessary.
[It is instead hypothetical] — Leontiskos
The most basic objection is that an argument with two conditional premises should not be able to draw a simple or singular conclusion (because there is no simple claim among the premises). — Leontiskos
[Using (b∧¬b) within formulas] is a bit like putting ethanol fuel in your gasoline engine and hoping that it still runs. — Leontiskos
as Tones' claim demonstrates. — Leontiskos
serious philosophical engagement — Leontiskos
Assertions, assertions, and more assertions. — Leontiskos
I want to say that (B∧¬B) and (B∧C) are not meta-logically equivalent, and because of this the truth tables are misleading. — Leontiskos
in everyday reasoning the truth of (3) requires the falsity of (1), even though P→~Q does not entail ~(P→Q), which indeed does seem to be a problem for material implication. — Srap Tasmaner
all, if P requires that ~Q, it can hardly require that Q. — Srap Tasmaner
people do recognize the difference even in everyday reasoning, and would accept that (2) is the simple contradiction of (1) — Srap Tasmaner
and that (3), while also denying (1) a fortiori, is a much stronger claim. — Srap Tasmaner
But when we place a contradiction in the consequent of a conditional it is no longer conditional (e.g. (A→(B∧¬B)). So if it is a meta-principle of classical propositional logic that all conditionals are conditional, then allowing the contradiction has upended this meta-principle. — Leontiskos
RAA pertains to the boundary of the system, not the interior. — Leontiskos
[...] what it means for a formula to be invalid. — Leontiskos
(A∧C)↔C is invalid for any (non-A) substitution of C. That's just what it means for a formula to be invalid. Yet when we substitute (B∧¬B) for C it magically becomes valid. — Leontiskos
One of the goals of the puzzle was to state the riddle in logic. Gödel's ontological proof has renditions of logical statements into plain English. The logical argument has been verified as valid, so we know that the plain English argument is valid too.I was reading various riddles and puzzles on the internet. One of the formulations of the riddles state: formulate it [the riddle] with first-order logic. — javi2541997
On the other hand, in English, or most European languages, nobody ever says "X implies false/true", that comes off as gibberish. — Lionino
I think the matter of bringing logical propositions into English and vice-versa is still quite meaningful. — Lionino
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