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Math Help - How is it true? p <=> p v p<=> (~p=>p).

  1. #1
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    How is it true? p <=> p v p<=> (~p=>p).

    Hello everybody.

    We have p <=> p v p<=> (~p=>p).
    I don't understand it. For example assume p: (2+2=4), so by this logical statements we have:

    (2+2=4)=> {if (2+2 is not equal to 4) then (2+2=4)}.

    and

    {if (2+2 is not equal to 4) then (2+2=4)}=> (2+2=4)

    Please help understand this problem.
    Thanks.
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    Quote Originally Posted by Researcher View Post

    p <=> p v p<=>(~p=>p).
    The above should be ~(~p=>p). It's the same as saying, "It is false that {If 2+2 \not=4, then 2+2=4.}"

    If p is true, then p v p is true. In logical expression p \Rightarrow p \vee p

    The conjunction is true if at least one is true.

    If the p v p is true, then p is true. In logical expression p \vee p \Rightarrow p

    (p \Rightarrow p \vee p) \vee (p \vee p \Rightarrow p) is equivalent to p \Leftrightarrow p \vee p.

    Now p \vee p \Leftrightarrow \sim \sim p \vee \sim \sim p.

    De Morgan's Law says \sim \sim p \vee \sim \sim p\Leftrightarrow \sim (\sim p \wedge  \sim p)

    \sim (\sim p \wedge  \sim p) \Leftrightarrow  \sim (\sim p \Rightarrow p)

    ---
    The conjunction p \vee p is like you are being offered to have a potato or a potato. There is only one choice, potato.

    It's different if the conjuction is p \vee q, which is like being offered rice or potato. If it's rice then it's not potato. So either way, you get at least one.

    p \Rightarrow p \vee p is like the same as saying, "If it's a potato, then you get potato or potato."

    Here is the De Morgan's Law \sim(P \vee Q) \equiv \sim P \wedge Q.

    Here is the negation of implication: \sim (P \Rightarrow Q) \equiv P \wedge \sim Q
    Last edited by novice; March 13th 2010 at 08:05 PM.
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  3. #3
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    Question

    Quote Originally Posted by novice View Post
    The above should be ~(~p=>p). It's the same as saying, "It is false that {If 2+2 \not=4, then 2+2=4.}"

    If p is true, then p v p is true. In logical expression p \Rightarrow p \vee p

    The conjunction is true if at least one is true.

    If the p v p is true, then p is true. In logical expression p \vee p \Rightarrow p

    (p \Rightarrow p \vee p) \vee (p \vee p \Rightarrow p) is equivalent to p \Leftrightarrow p \vee p.

    Now p \vee p \Leftrightarrow \sim \sim p \vee \sim \sim p.

    De Morgan's Law says \sim \sim p \vee \sim \sim p\Leftrightarrow \sim (\sim p \wedge  \sim p)

    \sim (\sim p \wedge  \sim p) \Leftrightarrow  \sim (\sim p \Rightarrow p)

    ---
    The conjunction p \vee p is like you are being offered to have a potato or a potato. There is only one choice, potato.

    It's different if the conjuction is p \vee q, which is like being offered rice or potato. If it's rice then it's not potato. So either way, you get at least one.

    p \Rightarrow p \vee p is like the same as saying, "If it's a potato, then you get potato or potato."

    Here is the De Morgan's Law \sim(P \vee Q) \equiv \sim P \wedge Q.

    Here is the negation of implication: \sim (P \Rightarrow Q) \equiv P \wedge \sim Q
    Thanks.

    We have; p \Leftrightarrow p \vee p
    and
    (p \vee q) \Leftrightarrow (\sim p \Rightarrow q) (by the law of contraposition)

    So if choosing p instead of q, we have;
    p \Leftrightarrow p \vee p \Leftrightarrow (\sim p \Rightarrow q)
    So what's wrong now?
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    Quote Originally Posted by Researcher View Post
    Thanks.

    We have; p \Leftrightarrow p \vee p
    and
    (p \vee q) \Leftrightarrow (\sim p \Rightarrow q) (by the law of contraposition) <---this is wrong.

    So if choosing p instead of q, we have;
    p \Leftrightarrow p \vee p \Leftrightarrow (\sim p \Rightarrow q)
    So what's wrong now?
    A contrapositive statement is the converse of an implication.

    If the implication is p \Rightarrow q, the contrapositive is q \Rightarrow p

    In other words, only an implication would have a contrapositive statement.
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  5. #5
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    Quote Originally Posted by novice View Post
    A contrapositive statement is the converse of an implication.
    If the implication is p \Rightarrow q, the contrapositive is q \Rightarrow p
    In other words, only an implication would have a contrapositive statement.
    That is wrong.
    The converse of p \Rightarrow q is q \Rightarrow p.
    The contrapositive of p \Rightarrow q is <br />
\neg q \Rightarrow \neg p.

    It is the case that \left[ {p \Rightarrow q} \right] \equiv \left[ {\neg p \vee q} \right]\,\& \,\neg \left[ {p \Rightarrow q} \right] \equiv \left[ {p \wedge \neg q} \right]
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    Quote Originally Posted by Plato View Post
    That is wrong.
    The converse of p \Rightarrow q is q \Rightarrow p.
    The contrapositive of p \Rightarrow q is <br />
\neg q \Rightarrow \neg p.

    It is the case that \left[ {p \Rightarrow q} \right] \equiv \left[ {\neg p \vee q} \right]\,\& \,\neg \left[ {p \Rightarrow q} \right] \equiv \left[ {p \wedge \neg q} \right]
    Yes, I have been very careless.

    The converse is q \Rightarrow p, but the contrapositive is \sim q \Rightarrow \sim p.

    Thank for pointing to it. I have to weed out my bad habit.
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  7. #7
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    Quote Originally Posted by Researcher View Post
    We have p <=> p v p<=> (~p=>p).
    I hope this formula is not parsed as (p\Leftrightarrow p)\lor(p\Leftrightarrow(\sim p\Rightarrow p)), because otherwise I have to think long and hard to decide whether this is true!

    Quote Originally Posted by novice
    The above should be ~(~p=>p).
    No, it is correct as stated in the original post. For all p and q we have (p\Rightarrow q)\Leftrightarrow(\sim p\lor q), so \sim p\Rightarrow p is equivalent to \sim\sim p\lor p, or p\lor p. Maybe the mistake was here:
    (outside negation has to be removed from either the left- or the right-hand side).
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  8. #8
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    Thumbs down

    Thanks friends.
    So i am right!
    Then, could you please help me to save me from this ambiguous issue?!
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    Quote Originally Posted by emakarov View Post
    ...outside negation has to be removed from either the left- or the right-hand side).
    Can't picture it in my head.

    P \wedge \sim Q \equiv \sim (P \Rightarrow Q)

    Now \sim P \wedge \sim Q \equiv \sim (\sim P \Rightarrow Q)

    Then \sim (\sim P \wedge \sim Q) \equiv \sim \sim (\sim P \Rightarrow Q)

    Substituting P for Q, we obtain

    \sim (\sim P \wedge \sim P) \equiv (\sim P \Rightarrow P)

    You are right. Did you do that in your head?

    Let P be the statement: Pigs fly.
    \sim P \Rightarrow P: If pigs don't fly, then pigs fly. Oh, lovely.

    I know, I know, F \Rightarrow T is T. That's why in math, it's called Trivial Proof, a useless proof.

    F \Rightarrow F is true; it's called Vacuous Proof, another useless proof.
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  10. #10
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    Angry

    So, don't you think it is a dubious logical expression?!
    I'm really getting crazy with this!!
    Please Help me to understand it ......
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    Quote Originally Posted by Researcher View Post
    So, don't you think it is a dubious logical expression?!
    I'm really getting crazy with this!!
    Please Help me to understand it ......
    No friend, they are no dubious logical. They have survived the test since ~350 B.C.

    I have given you all to help you understand.

    Beyond this, you need to see you teacher.

    Best wish.
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  12. #12
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    Thumbs up

    However i didn't understand it, a appreciate you for your helps!
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  13. #13
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    i think possibly that it may help if you stopped attempting to put real world examples to your expression, as you are studying classical logic (im assuming), it does not fit perfectly to real world solutions, i think instead you should try to understand it semantically using whatever methods of proof you have learned (e.g. tableaux, sequent calculus, resolution.......)
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