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Thread: Negation

  1. #1
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    Negation

    Hi,

    if $\displaystyle y \in f( \mathbf{A} \cup \mathbf{B}) \Leftrightarrow \exists x \in \mathbf{A} \text{ or } \exists x \in \mathbf{B} \text{ such as }y=f(x)$

    $\displaystyle \neg(y \in f( \mathbf{A} \cup \mathbf{B}))= y \not \in f( \mathbf{A} \cup \mathbf{B}))\Leftrightarrow \forall x \not \in \mathbf{A} \text{ and } \forall x \not \in \mathbf{B} \text{ we have }y=f(x)$

    it seems right to me but the last part seems weird "we have"...

    thanks in advance!
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  2. #2
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    Quote Originally Posted by sunmalus View Post
    $\displaystyle \forall x \not \in \mathbf{A} \text{ and } \forall x \not \in \mathbf{B} \text{ we have }y=f(x)$
    That's not right. It should be:

    $\displaystyle \forall x \in \mathbf{A} \text{ and } \forall x \in \mathbf{B} \text{ it is not the case that }y=f(x)$
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  3. #3
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    it's what I got at first but then I thought about it and the fact that
    $\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B}))=y \not \in f(\mathbf{A} \cup \mathbf{B})$
    made me think that is was wrong... So last question is that negation right or is it $\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B})=y \not \in f(\mathbf{A} \cap \mathbf{B})$??

    logically I'd say
    $\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B}))=y \not \in f(\mathbf{A} \cup \mathbf{B})$
    but once again it seems weird...
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  4. #4
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    Quote Originally Posted by sunmalus View Post
    $\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B}))=y \not \in f(\mathbf{A} \cup \mathbf{B})$
    No, it should be:

    $\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B})) \Leftrightarrow y \not \in f(\mathbf{A} \cup \mathbf{B})$

    In this kind of context, an equal sign doesn't go between two statements. Rather, to express that two statements are equivalent, use the biconditional sign.

    Or (and this is a picky technical point having not much substantive to do with the subject of images that these formulas are about), you could say:

    "$\displaystyle y \not \in f(\mathbf{A} \cup \mathbf{B})$" stands for "$\displaystyle \neg$ $\displaystyle y\in f(\mathbf{A} \cup \mathbf{B})$".

    Anyway, all you need is:

    $\displaystyle \neg$ $\displaystyle y\in f(\mathbf{A} \cup \mathbf{B}) \Leftrightarrow y \not \in f(\mathbf{A} \cup \mathbf{B})$
    Last edited by MoeBlee; Oct 20th 2010 at 08:42 AM.
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  5. #5
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    Darn, I needed to edit my previous post because I messed up MATH tags. Please read post again now edited.

    Anyway, without messing with tags, in plain ASCII:

    ~ y in f[A u B] <-> y not in f[A u B]
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  6. #6
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    Last questions:
    $\displaystyle \forall x \in \mathbf{A} \cap \forall x \in \mathbf{B}$

    is not the same as:

    $\displaystyle \forall x \in \mathbf{A} \cap \mathbf{B}$

    right?

    If not what is another way to represent
    $\displaystyle \forall x \in \mathbf{A} \cap \forall x \in \mathbf{B}$

    (I think this was my real problem the whole time)
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  7. #7
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    Quote Originally Posted by sunmalus View Post
    $\displaystyle \forall x \in \mathbf{A} \cap \forall x \in \mathbf{B}$
    That makes no sense. It is not well formed, not syntactically correct.

    Probably what you mean is this:

    $\displaystyle \forall$x((x$\displaystyle \in$A & x$\displaystyle \in$B) -> x $\displaystyle \in$A$\displaystyle \cap$B)

    Or even just leave off the initial universal quantifier and let it be understood as tacit:

    (x$\displaystyle \in$A & x$\displaystyle \in$B) -> x $\displaystyle \in$A$\displaystyle \cap$B

    I think you need to review basic symbolic logic to see how formulas are formed. I would suggest 'Logic: Techniques Of Formal Reasoning' by Kalish, Montague, and Mar.
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  8. #8
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    $\displaystyle \neg y \in f( \mathbf{A} \cup \mathbf{B})\underbrace{=}_{\text{this should be} \Leftrightarrow} y \not \in f( \mathbf{A} \cup \mathbf{B}))\Leftrightarrow \underbrace{\forall x \in \mathbf{A} \text{ and } \forall x \in \mathbf{B}}_{\not \Leftrightarrow \forall x \in \mathbf{A} \cap \mathbf{B}} \text{ is not the case that }y = f(x)$

    this is what I meant but I thought it was clear because of all the previous posts, sorry.
    Last edited by sunmalus; Oct 20th 2010 at 12:19 PM.
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  9. #9
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    It should be pointed out that $\displaystyle \forall x \in \mathbf{A} \text{ and } \forall x \in \mathbf{B} \text{ it is not the case that }y=f(x)$ is not a well-formed proposition (formula). The inductive definition of formulas goes like this. "... If P and Q are formulas, then $\displaystyle P\land Q$ is a formula... If P(x) is a formula that may contain a free variable x, then $\displaystyle \forall x.\,P(x)$ is a formula." So something that starts with $\displaystyle \forall x\in A\land{}$ is not a well-formed formula. (The culprit, of course, is not $\displaystyle \in A$.)

    If you write the statement as

    $\displaystyle (\forall x\in A.\,y\ne f(x))\land(\forall x\in B.\,y\ne f(x))$

    reasoning about it would be easier.
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  10. #10
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    No, you repeated a previous error.

    (I'm tired of messing with LaTex tags, so I'm going to give you what you want in plain ASCII

    ~ stands for negation
    e stands for membership
    <-> will stands for material equivalence
    u stands for binary union
    U stands the universal quantification
    v stands for disjunction
    -> stands for material implication
    not= stands for inequality)

    ~ y e f[AuB] <-> y not in f[AuB] <-> Ux((x e A v x e B) -> ~ y=f(x)) <-> UxeA UxeB y not= f(x) <-> Ux(x in AuB -> ~ y=f(x))

    In other words:

    it is not the case that y is in f[AuB]

    if and only if

    y is not in f[AuB]

    if and only if

    for all x, if (x in A or x in B) then ~ y=f(x)

    if and only if

    for all x in A and all x in B, ~ y=f(x)

    if and only if

    for all x in AuB, ~ y=f(x)

    Take my suggestion to brush up on your symbolic logic?
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  11. #11
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    Thank to you all but special thanks to MoeBlee! I really get it now and yes I'll take your suggestion .
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  12. #12
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    delete post
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