arguments

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Dr.Peterson said:
You need to consider what "valid" means. You have given a correct definition.

Validity is only about the form of the argument, not its content (the truth or falsity of the statements within it). An argument can be invalid even if all its premises and its conclusion are true, as in this case.

Your argument has the form "If p, then q; not p; therefore not q." So you have to decide whether this is a valid argument, by considering whether it is possible for both premises to be true but the conclusion false. (The answer is that it is invalid; specifically, it is the fallacy of the inverse.)

I believe you have shown in the past that you know something about logic; so this should not be hard for you. Perhaps what you mean by "challenging question" is not "a question that is hard for me, for which I need help", but "a challenge question that I think you will get wrong." In either case, please tell us your thoughts about it.
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Challenging means I know the unswer and I challenge the others to find the solution to my question
Now consider your self having to answer that question in ancient Greece where symbolic logic had not been discovered yet, what would be your answer
Dr.Peterson said:
If I understand you correctly, you are demonstrating that the argument in the OP is invalid (as I said in #6, it's the fallacy of the inverse), by showing an instance of the same argument in which the premises are true but the conclusion is false.

Premise 1 is true because its condition is false; premise 2 is true; but the conclusion is false.

I agree.

I suspect people are thinking you are disagreeing with them, when you are not.
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Correct
AvgStudent said:
Thank you, @JeffM and @Dr.Peterson, for the insights.
This was a great explanation and showed just how little I know about logic. I'm definitely intrigued to learn more about logic and arguments.
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I wonder
Should not logic be introduced
And studied in high shcool
Logic is the study of arguments
The questions that comes next are
1 what is an argument
2 How do we find arguments in our speech written or oral
3 when is argument valid or non valid
And from here onwards we have the split of logic into
FORMAL AND INFORMAL e. t. c e. t. c
By theway

By the way a mathematical proof is none other than an argument
The laws of logic are behind any mathematical argument (mathematical proof),physical chemical ,legal argument
lex said:
Valid argument [imath]\Leftrightarrow[/imath] (True premises [imath]\Rightarrow[/imath] True Conclusion)
This is equivalent to:
Invalid argument [imath]\Leftrightarrow[/imath] Not(True premises [imath]\Rightarrow[/imath] True Conclusion)
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CAN you prove that in statment calcules in other words can you prove:

[math][P\Leftrightarrow Q]\Leftrightarrow[\neg P\Leftrightarrow \neg Q][/math]Valid argument =P
And Q= (True premises [imath]\Rightarrow[/imath] True Conclusion)

I may point out again that true tables can ensure us that the above is provable but they do not provide a proof

Sorry I answered to your post No 6 before your answer no 22
 
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The point is that the notion that [imath]A \land B[/imath] is logically equivalent to [imath]\lnot(A \rightarrow \lnot B)[/imath] is a notion outside of the formal system L (just as my 'informal' argument was). [imath]\land[/imath] is not part of L and is not defined there.
 
can you prove then in your system that : ~(A->~B)->A
 
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Yes!
As you can see, constructing proofs in L is rather 'long-winded' and tedious.
Thankfully there is a theorem about L, called the "Adequacy Theorem", which shows that if a wff in L is a tautology then the wff is a theorem of L. We can decide whether a wff is a tautology by examining the truth table of the corresponding statement.
L is thus 'decidable' and this theorem allows us to avoid constructing proofs in L.
 
So if you sustitute in the above formula ~(A->~B) WITH A&B dont you get A&B-> A? and {A,B}-> ~(A->~B)<->A&B?
Dont you define in your system A<->B) as (A-.B)&(B->A) AS you have shown in your post No 31?
 
chrislav said:
So if you sustitute in the above formula ~(A->~B) WITH A&B dont you get A&B-> A? and {A,B}-> ~(A->~B)<->A&B?
Dont you define in your system A<->B) as (A-.B)&(B->A) AS you have shown in your post No 31?
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So if you sustitute in the above formula ~(A->~B) WITH A&B dont you get A&B-> A?
If we are talking about formal statement calculus L, [imath](A \land B) \rightarrow A[/imath] is not a wff in L, as [imath]\land[/imath] is not in the alphabet of L.

{A,B}-> ~(A->~B)<->A&B
If you are saying [imath]\{A,B\} \underset{L}{\vdash} \lnot (A \rightarrow \lnot B) \leftrightarrow (A \land B)[/imath] then this is not correct, because again [imath]\land[/imath] is not in the alphabet of L.

Dont you define in your system A<->B) as (A-.B)&(B->A) AS you have shown in your post No 31?
[imath]A\leftrightarrow B[/imath] does not exist in L as [imath]\leftrightarrow[/imath] is not in the alphabet of L (and neither is [imath]\land[/imath]).
What I was doing in #31 was trying to interpret outside of L, your statement in #26:
[math][P\Leftrightarrow Q]\Leftrightarrow[\neg P\Leftrightarrow \neg Q][/math]I interpreted [imath]A \leftrightarrow B \text{ as } A\rightarrow B, B\rightarrow A[/imath], each of which exists in L.
In post #26 you say:
chrislav said:
CAN you prove that in statment calcules in other words can you prove:

[P⇔Q]⇔[¬P⇔¬Q]
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The simple answer is No, as the expression does not exist in L.
I was trying to be helpful in interpreting it as expressions which do exist in L and then giving an example of a proof in L.

Of course there is a 'logically equivalent' form of your statement, which exists in L, and is a theorem in L.
I repeat that we have two ways of proving that something is a theorem in the formal system L:
(i) Construct a proof in L
(ii) Show that it is a tautology, by checking the corresponding statement's truth table and then invoking the Adequacy Theorem.
Either of these methods proves that the L-version of the statement is a theorem in L.
 
So what is it
Can you or cannot you prove the above equivalence in any deduction system you wish to choose?
 
chrislav said:
So what is it
Can you or cannot you prove the above equivalence in any deduction system you wish to choose?
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We do not get involved in useless diatribe. Find some other goat, some other herd and prance around there....
 
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Subhotosh Khan said:
We do not get involved in useless diatribe. Find some other goat, some other herd and prance around there....
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proving A<->B)<->(~A<->~B) is a diacritic problem in propositional calculus and not useless diatribe
For example it is mention as a problem in the book" logic" of :schaum's series page 70
 
chrislav said:
So what is it
Can you or cannot you prove the above equivalence in any deduction system you wish to choose?
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Please refer to my previous answers.
I am inclined to agree with Subhotosh Khan that this thread may have run its course.
 
here is a mathematical forum and nobody up to now has produced a proof of the formula (A<->B)<->(~A<->~B) in any system in propositional calculus whether axiomatic or natural.
And iI challenge the whole forum to give a proof
 
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