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Math Help - de morgan's laws

  1. #1
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    de morgan's laws

    I was looking through my old notes, trying to refresh my memory and I can't prove ~(A ^ B) -> (~A v ~B). Can anyone help? The rules I'm letting myself use are conjunctive addition, simplification, disjunctive addition, disjunctive syllogism, modus ponens, double negation. I can use indirect proof and conditional proof.
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    Quote Originally Posted by jdc4 View Post
    I was looking through my old notes, trying to refresh my memory and I can't prove ~(A ^ B) -> (~A v ~B). Can anyone help? The rules I'm letting myself use are conjunctive addition, simplification, disjunctive addition, disjunctive syllogism, modus ponens, double negation. I can use indirect proof and conditional proof.
    You should not be asking for us to give a tutorial is logic. That is not what the forum is about.
    If you have a specific question, then post it. We will try to help.

    Having said that, here is some remarks on your first question.
    A \wedge B is true if and only if both A & B are true.
    So if we negate that statement, \neg \left( {A \wedge B} \right), then that means that at least one of the conjuncts is false.
    That is \neg A \vee \neg B.
    The converse relies upon \neg (\neg A) \equiv A.
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    Hi Plato, I wasn't asking for a tutorial. I was asking for someone to help me prove that theorem. I apologize if I wasn't clear. Thanks for the help, but I can tell that it's valid. I want to prove it with those rules and I can't seem to do it tho I'm sure it can be done. Thats what I need help with.
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  4. #4
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    Quote Originally Posted by jdc4 View Post
    Hi Plato, I wasn't asking for a tutorial. I was asking for someone to help me prove that theorem. I apologize if I wasn't clear. Thanks for the help, but I can tell that it's valid. I want to prove it with those rules and I can't seem to do it tho I'm sure it can be done. Thats what I need help with.
    Then I must ask you: "What Rules?"
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    Modus ponens: from P and P -> Q, infer Q l Conjunctive addition: from P and Q, infer (P ^ Q) l Simplification: from (P ^ Q), infer P; from (P ^ Q) infer Q l Disjunctive addition: from P, infer (P v Q); from P, infer (Q v P) l Disjunctive syllogism: from (P v Q) and ~ P, infer Q; from (P v Q) and ~Q infer P l Double negation: from P, infer ~~P I can use both conditional proof and indirect proof.
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  6. #6
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    Quote Originally Posted by jdc4 View Post
    Modus ponens: from P and P -> Q, infer Q l Conjunctive addition: from P and Q, infer (P ^ Q) l Simplification: from (P ^ Q), infer P; from (P ^ Q) infer Q l Disjunctive addition: from P, infer (P v Q); from P, infer (Q v P) l Disjunctive syllogism: from (P v Q) and ~ P, infer Q; from (P v Q) and ~Q infer P l Double negation: from P, infer ~~P I can use both conditional proof and indirect proof.
    I am sorry to say, "What is the point of all of that?"
    It seems that you know that. I certainly know it, having taught Symbolic Logic for years.
    So what is you point or your question?
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  7. #7
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    Well I tried proving it with those rules and I couldn't, and now it's kind of an obsession. It's hard to explain. When I try solving a problem and I can't, I want to know how it can be done. I'm sure it's actually very simple but for some reason I'm not getting it. I'd be surprised if it took more than 7-10 lines.
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  8. #8
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    Quote Originally Posted by jdc4 View Post
    I was looking through my old notes, trying to refresh my memory and I can't prove ~(A ^ B) -> (~A v ~B). Can anyone help? The rules I'm letting myself use are conjunctive addition, simplification, disjunctive addition, disjunctive syllogism, modus ponens, double negation. I can use indirect proof and conditional proof.

    Assume ~(A^B)............................................ .............................1

    Assume A................................................. .................................2

    Assume B for contradiction..................................... ........................3


    A & B .....from (1) and (2) and using conjunctive addition , or differently called addition introduction...................................... .............................4


    ~(A^B) & (A^B) ......from (1) and (4) and using again conjunctive addition.......................................... .......................................5

    ~B...... BY applying contradiction from (3) to (5)............................6


    A------>~B......BY applying the conditional proof rule from (2) to (6)............................................... ................................................7



    Now by material implication rule ( pvq <-----> ~p---->q) we can convert



    A------>~B INTO ~AV~B
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