If the OP uses propositional logic, it doesn't use propositional logic.
It uses prepositional logic
Therefore it doesn't use propositional logic.
MP has spoken. It doesn't use prepositional logic — Hanover
But (A->~A) & A is a contradiction.
If you assert A->~A, and then go on to assert A, then you have contradicted yourself. — Srap Tasmaner
A->A, which is either useless — Srap Tasmaner
it seems to me letting mathematical logic have the last word is the tail wagging the dog. — Srap Tasmaner
Well for a start you would no longer be dealing with a complete version of propositional calculus... — Banno
Do you, Srap, agree that the argument in the OP is valid? — Banno
I cannot think of a way to frame this as a real example — Count Timothy von Icarus
I get that MP requires a "->", but (A->~A)<->~A, so I'm puzzled by insisting on this nicety. In classical logic it's materially equivalent to the disjunctive syllogism, isn't it? — Srap Tasmaner
1. If Hanover is correct, Hanover is not correct
2. Hanover is correct
3. Hanover is not correct (1,2 mp)
4. Hanover is not correct or 3 is an invalid conclusion derived from mp.(3, introduction) — Hanover
1. Meaning what exactly? — Srap Tasmaner
2. Is the answer to (1) something I should care about? — Srap Tasmaner
Do you, Srap, agree that the argument in the OP is valid?
— Banno
I don't really care. It's abusive. — Srap Tasmaner
Exactly that. If you modify the substitution rule to remove substitution of the same variable on both sides of a function, can you demonstrate that the resulting calculus will be complete? Can you prove A→A, for example?1. Meaning what exactly? — Srap Tasmaner
That's entirely up to you. But you are on this thread, so forgive my presumption. Failing to see that the argument in the OP is valid is an indication of a lack of understanding of basic logic. Refusing to give an opinion says something else.Is the answer to (1) something I should care about? — Srap Tasmaner
Some folk think that pointing out an error os abusive. Odd, sad, but trueBut what do you mean by 'abusive'? — TonesInDeepFreeze
to claim "A → ~A" along with "A" is to contradict oneself, and therein I think lies the confusion. — Count Timothy von Icarus
Modus ponendo ponens is the principle that, if a conditional holds and also its antecedent, then its consequent holds." (Beginning Logic - Lemmon)
Perhaps your argument is based on taking that to mean this?:
If a conditional holds and also its antecedent, then modus ponedo ponens is the principle that then its consequent holds.
— TonesInDeepFreeze
Be good enough to make clear the difference between these two. — tim wood
But I can't say what is the source of the mental block in people who don't understand that "A -> ~A with A is contradictory
But I can't say what is the source of the mental block in people who don't understand that "A -> ~A with A is contradictory
But I can't say what is the source of the mental block in people who don't understand that "A -> ~A with A is contradictory" doesn't entail "A -> ~A is contradictory" — TonesInDeepFreeze
1. A -> not-A
2. A
Therefore,
3. not-A.
— NotAristotle
#1 is a contradiction, reducible to ~ A or ~A. — Hanover
Well, if you've been taught that a contradiction has a truth table that is always false and you think you have identified something that is necessarily/always false, it seems possible to conflate the two. — Count Timothy von Icarus
mental buckets — Count Timothy von Icarus
can you demonstrate that the resulting calculus will be complete? — Banno
If you are interested in the basics of ordinary formal logic, then it would be a question that would naturally occur to you. But I don't see why you couldn't study other branches of philosophy without understanding the completeness of the propositional calculus. — TonesInDeepFreeze
Can you prove A→A, for example? — Banno
But what do you mean by 'abusive'? — TonesInDeepFreeze
3 follows from 1 and 2 by modus ponens.
— TonesInDeepFreeze
1 means "If A is true, A is false." This means A can never be true, despite it being true. It's a walking contradiction. This in itself can be taken to mean A is false because, as noted A -> ~A is logically equivalent to ~A or ~A as a disjunction of the conditional ( A --> B = ~A or B). 1 therefore means ~A.
This can be reduced to:
1. ~ A
2. A
Therefore ~A.
The conclusion is a restatement of #1. 2 is a contradiction of 1.. — Hanover
There is, in this case, a veneer of logic over what could scarcely be considered rational argumentation. If this appearance of rationality serves any purpose, it must be to mislead, hence abusive, eristic, sophistical, non-cooperative.
But for the everyday use of logic just to schematize and clarify arguments, you get a lot more mileage out of de Morgan's laws, contrapositives, a solid understanding of quantifiers, and such. — Srap Tasmaner
But what do you mean by 'abusive'?
— TonesInDeepFreeze
The basic idea is "formally correct but misleading or akin to sophistry". Akin to sophistry. — Srap Tasmaner
People worry over the sense in which the conclusion of a deductive argument is "contained" in the premises ― here it is one of the premises. — Srap Tasmaner
In what sense is the relationship of A and ~A revealed or clarified? It may be modus ponens in form, but hardly in spirit — Srap Tasmaner
And if we step back and look at the offending premise, we get to ~A by noting that A→~A is materially equivalent to ~A v ~A. Now what kind of disjunction is that? It's a well-formed-formula ― no one can deny that ― but it's hardly what we usually have in mind as a disjunction. It's "heads I win, tails you lose." That's abusive. — Srap Tasmaner
There is, in this case, a veneer of logic over what could scarcely be considered rational argumentation. If this appearance of rationality serves any purpose, it must be to mislead, hence abusive, eristic, sophistical, non-cooperative. ― Again, I am only talking about how logic is used as an aid to ordinary philosophizing, not what people get up to in a logic lab. — Srap Tasmaner
There is, in this case, a veneer of logic over what could scarcely be considered rational argumentation. If this appearance of rationality serves any purpose, it must be to mislead, hence abusive, eristic, sophistical, non-cooperative.
Heh, describes how these threads normally go. — Count Timothy von Icarus
A --> B = ~ A v B.
A --> ~A = ~A v ~A
~A v ~ A = ~A — Hanover
Then you'd argue incorrectly
— TonesInDeepFreeze
This is where we disgree.
A --> ~A <> A --> ~ B because A-->~A = ~A, yet A-->~B <> ~A. — Hanover
In this case, for instance, it is suggested that we conclude ~A by modus ponens. — Srap Tasmaner
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