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Math Help - logic question

  1. #16
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    Quote Originally Posted by Jhevon View Post
    no, this argument is not valid. .
    But you also wrote and i quote:

    "you were given an argument to prove its validity. that simply means you want to show that the conclusion follows logically from the premises. whether the premises are true or not is immaterial, your job is to show that IF they are true, the conclusion would have to be true."


    And according to that the argument is valid ,since, if we assumed the premises to be true then the conclusion is true as i showed
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  2. #17
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by archidi View Post
    But you also wrote and i quote:

    "you were given an argument to prove its validity. that simply means you want to show that the conclusion follows logically from the premises. whether the premises are true or not is immaterial, your job is to show that IF they are true, the conclusion would have to be true."


    And according to that the argument is valid ,since, if we assumed the premises to be true then the conclusion is true as i showed
    as i said, you begged the question. you did not show it. the conclusion does not follow, even if the premises were true. it is an invalid argument, any proof you give will be wrong.

    the argument:

    (1) P => Q
    (2) Q
    ------------
    P

    is invalid, the conclusion does not follow logically from the premises, EVEN IF the premises are true.

    you can't just do what you want and say what you want to come up with the conclusion. there are rules that you have to follow. of course when i say "follow logically from the premises" i mean it can be "derived from the premises using the rules of inference for logic"


    please review the laws of logic... did you take my advice and looked up what "valid" and "sound" means?
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  3. #18
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    I have defined validity at the beginning of my post #11.

    an argument is valid if it is provable,syntactically.

    In propositional calculus an argument is valid iff it is a theorem iff it is a tautology
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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by archidi View Post
    I have defined validity at the beginning of my post #11.
    in post #11, you stated what it meant for a particular kind of argument (namely, one with 4 premises, which may or may not have anything to do with each other) to be valid, not what it means for arguments in general to be valid


    an argument is valid if it is provable,syntactically.

    In propositional calculus an argument is valid iff it is a theorem iff it is a tautology
    strange definitions.

    nevertheless, your proof fails in both areas, so your argument is invalid in any case.

    as i said, we can show that it is not a tautology via truth tables. you would have to be more specific as to what you mean by "provable syntactically" but i suppose it means what i was saying, "provable by the laws of logic." in that case, the proof does not work either.

    we are really going around in circles here. the proofs that Soroban and I gave are ok. so far, the proofs you have given are not. apparently you do not like our proofs, i can't help you there. but what you are doing is not very productive. either take one of the proofs and figure out why it works, or wait on your professor to give a solution, hopefully more to your liking. i do suggest that you re-read your text and think carefully about what it says, and try to get the concepts straight. make sure you know what premises are and how to treat them. what is means to prove, what validity entails, etc. it might seem straight forward, but sometimes these things are not. in general, logic does not follow the way most people think naturally, at least, it is a lot more strict on how it goes from one thought to another.
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  5. #20
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    Quote Originally Posted by archidi View Post

    an argument is valid if it is provable,syntactically.

    In propositional calculus an argument is valid iff it is a theorem iff it is a tautology
    Quote Originally Posted by Jhevon View Post


    strange definitions.



    you would have to be more specific as to what you mean by "provable syntactically" but i suppose it means what i was saying, "provable by the laws of logic." in that case, the proof does not work either.

    The 2nd of the two sentences i wrote and in particular: "an argument is valid iff it is a theorem iff it is a tautology".


    It is not a definition but the central theorem in propositional calculus.


    It is the theorem joining the semantical part with the syntactical part of the propositional calculus,and at the same time proving the completeness of propositional logic.

    Propositional calculus is separated in two parts :

    1) The semantical part ,in which every valid argument is thru the true tables proved to be a tautology.


    2)The syntactical part ,where every valid argument is proved to be a theorem with the help of other basic theorem (called the laws of logic or rules of inference). These basic theorems result from a set of axioms,the axioms of propositional calculus.

    The proof of a valid argument in the 1st part is called semantical proof ,while its proof within the 2nd part is called syntactical proof.


    The two parts are joined together by the theorem :

    Every valid argument is a theorem and every theorem is a tautology.


    A valid argument =====> is a theorem ( in the syntactical part of propositional logic) ======> is a tautology.

    The reverse of that theorem:

    Every tautology is a theorem and every theorem is a valid argument proves the completeness of propositional logic ,provided that propositional calculus is defined to be the set of valid arguments.

    The two above theorems are joined together to form,the famous central theorem of propositional logic:

    An argument is valid iff (<======>) it is a theorem iff(<====> ) it is a tautology
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  6. #21
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    Soroban ,in your proof you have assumed that :

    ~p=~(1>0) ,also ~p= 1\leq 0.


    Hence ,when you prove.... ~p... not only you prove  1\leq 0 but also ~(1>0).

    Therefor we have a proof with two answers.

    But this is not possible
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