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Math Help - Please help!

  1. #1
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    Question Please help!

    How to prove that this below argument is valid, using established Rules of reasoning and Logical Equivalences (using vertical statement -reason format.)

    {[(p ˅ q ) → r ] ˄ [ r →(s v t)] ˄ (s ˄ u)˄(u → t)}→ p ??????




    p.s For those who don't see the symbols I will put this argument in words:
    {[( not p v q) --> r] and [r --> (s v t)] and (not s and not u) and ( not u --> not t)} -->p
    Last edited by olenka; September 18th 2008 at 03:17 PM. Reason: cannot see the symbols
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  2. #2
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    Ummm - keep simplifying using
    a -> b = !a + b
    !(a + b) = !a * !b
    !(ab) = !a + !b
    where ! means "not", + means "or", and * or multiplication means "and".

    To prove that
    {[( not p v q) --> r] and [r --> (s v t)] and (not s and not u) and ( not u --> not t)} -->p
    = ((!p + q) -> r)(r -> (s+t))(!s)(!u)(!u -> !t)) -> p
    = (p(!q)+r)(!r+s+t)(!s)(!u)(u+!t)) -> p
    Now (!u)(u+!t) = !u !t
    (!r+s+t)(!s)(!t) = !r !s !t
    So LHS = (p !q + r)(!r)(!s)(!t)(!u) =
    p(!q)(!r)(!s)(!t)(!u)
    SO
    To prove that
    p(!q)(!r)(!s)(!t)(!u) -> p or
    !p + q + r + s + t + u + p
    which is true since it has !p + p
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  3. #3
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    Quote Originally Posted by olenka View Post
    How to prove that this below argument is valid, using established Rules of reasoning and Logical Equivalences (using vertical statement -reason format.)

    {[(p ˅ q ) → r ] ˄ [ r →(s v t)] ˄ (s ˄ u)˄(u → t)}→ p ??????




    p.s For those who don't see the symbols I will put this argument in words:
    {[( not p v q) --> r] and [r --> (s v t)] and (not s and not u) and ( not u --> not t)} -->p

    We want to prove that all the statements logically imply p

    We notice that the statement having p is the statement (~pvq)--->r but in the proof that it will follow the way it is written down it will not help us to get p out of it.So through contrapositive law we can convert it to:


    ........................~r-----> ~(~pvq)..........................................1

    Now we convert r---->(svt) by contrapositive law again to:

    .........................~(svt)----->~r............................................. .2

    we are also given as assumptions :

    ...........................~u..................... .........................................3

    ..............................~s.................. ............................................4

    From ....~u----> ~t and 3 and using M.Ponens we get :


    ...............................~t................. ............................................5

    From 4 and 5 and using the law of logic called addition introduction we get :

    ..............................~s^~t............... ..........................................6

    From ...6 and using the law called De Morgan we get:


    ................................~(svt)............ .............................................7

    Now from 2 and 7 and using the law called M.Ponens we get:

    .................................~r............... ...............................................8

    And from 1 and 8 and using again M.Ponens we get :

    .........................................~(~pvq )...........................................9

    From 9 and using again De Morgan we get :

    .........................................p^~ q..............................................10

    And finally from 10 and using addition elimination we get :


    ...........................................p...... ..............................................11


    NOTE the above logically implies ~q as well.

    The above can be written in amore tabular form, like the following:

    1) (~pvq)------>r................................................ ......assumption

    2) r----->(svt)............................................ ...............assumption

    3) ~s................................................ ........................assumption

    4) ~u................................................ .......................assumption

    5) ~u------>~t............................................... ...........assumption

    6) ~r-----~(~pvq).................................... 1, contra positive

    7) ~(svt)------>~r......................................2,contr a positive

    8) ~t................................................ .......4,5 M.Ponens

    9) ~s^~t............................................. ......3,8 addition intr.

    10) ~(svt)............................................ .....9, De Morgan

    11) ~r................................................ .......7,10 M.Ponens

    12) ~(~pvq)........................................... .....6,11 M.Ponens

    13) p^~q.............................................. .....12, De Morgan

    14) ............p..................................... .........13 addition elimination


    The laws use are:

    a) contra positive:........(P----->Q) ------>(~Q----->~P) and also (~Q------>~P)-------->(P------>Q)

    b) M.Ponens :..........(P----->Q AND P)------>Q

    c) De Mrgan...........~(P ^Q) <-------> ~P v ~Q

    d) addition introduction ...... P,Q --------> P^Q

    e) addition elimination......P^Q------>P

    THERE may be other shorter or longer proofs the above only shows the way

    Also as hwhelper mentioned there can be a Boolean proof by substituting "a---->b" with ~avb.

    Finally by the use of true tables there can be a semantical proof in which we must show that:


    The conditional:


    {[(p ˅ q ) → r ] ˄ [ r →(s v t)] ˄ (s ˄ u)˄(u → t)}→ p

    IS A tautology
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  4. #4
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    Talking

    Thank you so much guys ! U R the best!!!!
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  5. #5
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    Quote Originally Posted by olenka View Post
    Thank you so much guys ! U R the best
    I for one am not so sure about that. It is not standard.
    \begin{gathered}   1.\,\left( {\neg p \vee q} \right) \to r \hfill \\<br />
  2.\,r \to \left( {s \vee t} \right) \hfill \\  3.\,\neg s \wedge \neg u \hfill \\<br />
  4.\,\neg u \to \neg t \hfill \\  \therefore \,p \hfill \\ \end{gathered}

    \begin{gathered}<br />
  5.\,\left( {\neg p \vee q} \right) \to \left( {s \vee t} \right)\;\left[ {1\,\& \,2} \right] \hfill \\<br />
  6.\,\neg u\;\left[ 3 \right] \hfill \\<br />
  7.\,\neg t\;\left[ 4 ,\& \,6\right] \hfill \\<br />
  8.\,\neg s\;\left[ 3 \right] \hfill \\<br />
  9.\,\neg s \wedge \neg t\left[ {7\,\& \,8} \right] \hfill \\ <br />
\end{gathered}
    \begin{gathered}  10.\,\neg \left( {s \vee t} \right)\left[ 9 \right] \hfill \\  11.\,\neg \left( {\neg p \vee q} \right)\left[ {10\,\& \,5} \right] \hfill \\  12.\,p \wedge \neg q\left[ {11} \right] \hfill \\  \therefore \,p\left[ {12} \right] \hfill \\ \end{gathered}

    I will leave it to you to supply the reasons by name. Names of the rules differ from text to text and author to author. That is the reason I don’t trust what you have handed. I have taught this material for years in North America. I can tell you that I got confused each time we changed textbooks.
    Notations and rules names are simply not standard.

    However, the layout of the proof that I have given you is fairly standard.
    Last edited by Plato; September 19th 2008 at 03:45 PM.
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  6. #6
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    Here is ashorter proof by contradiction:

    Let assume ~p and start a contradiction,then by the law called disjunction introduction we have :

    ...........................~pvq................... ...........................................1

    from the assumption ~pvq----> r and 1 and using M.Ponens we get:


    ...................................r.............. ................................................2

    from the assumption r----->(svt) and 2 and using M.Ponens we get :


    .........................................(svt).... ..............................................3


    from the assumptions ~u------>~t and ~u and using M.Ponens we get:


    ...............................................~t. ...............................................4


    from the assumption ~s and 4 and using addition introduction we get:


    .......................................~s^~t...... ...........................................5

    from 5 and using De MORGAN we get :


    .....................................~(svt)....... ...........................................6


    from 3 and 6 and using addition introduction we get :



    ....................................(svt) and ~(svt)......................................7


    ...................................a contradiction..................................... ...


    HENCE .........................P........................ ..................................
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  7. #7
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    Quote Originally Posted by Plato View Post
    I for one am not so sure about that. It is not standard.
    \begin{gathered} 1.\,\left( {\neg p \vee q} \right) \to r \hfill \\<br />
2.\,r \to \left( {s \vee t} \right) \hfill \\ 3.\,\neg s \wedge \neg u \hfill \\<br />
4.\,\neg u \to \neg t \hfill \\ \therefore \,p \hfill \\ \end{gathered}

    \begin{gathered}<br />
5.\,\left( {\neg p \vee q} \right) \to \left( {s \vee t} \right)\;\left[ {1\,\& \,2} \right] \hfill \\<br />
6.\,\neg u\;\left[ 3 \right] \hfill \\<br />
7.\,\neg t\;\left[ 4 ,\& \,6\right] \hfill \\<br />
8.\,\neg s\;\left[ 3 \right] \hfill \\<br />
9.\,\neg s \wedge \neg t\left[ {7\,\& \,8} \right] \hfill \\ <br />
\end{gathered}
    \begin{gathered} 10.\,\neg \left( {s \vee t} \right)\left[ 9 \right] \hfill \\ 11.\,\neg \left( {\neg p \vee q} \right)\left[ {10\,\& \,5} \right] \hfill \\ 12.\,p \wedge \neg q\left[ {11} \right] \hfill \\ \therefore \,p\left[ {12} \right] \hfill \\ \end{gathered}

    I will leave it to you to supply the reasons by name. Names of the rules differ from text to text and author to author. That is the reason I dont trust what you have handed. I have taught this material for years in North America. I can tell you that I got confused each time we changed textbooks.
    Notations and rules names are simply not standard.

    However, the layout of the proof that I have given you is fairly standard.

    The names may vary from book to book but the structure of the law is still the same .

    for addition elimination for example you can write Q^P------> P e.t.c ,e.tc

    And since you taught logic i suppose you will be able to give us a stepwise proof like the above where the laws of logic are explicitly mentioned of the following:

    \forall x \forall y( x \geq 0 and y \geq 0--------> \sqrt{xy}= \sqrt{x} \sqrt{y})

    Also any theorems axioms definitions involved i.e justification of each step



    ......................................here and now............................................... .........................
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  8. #8
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    Quote Originally Posted by Plato View Post
    That is the reason I dont trust what you have handed. .


    IN propositional calculus you check each and every exercise if it s provable by using the true tables.

    If the exercise is a TAUTOLOGY then it is provable.


    Propositional calculus is


    ........................................ a DECIDABLE THEORY......................................

    SO if youdo not trust what she handed out simply go to the true tables and see if what she handed out is a Tautology or not.
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  9. #9
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    Triclino, If you had any idea of what is really going on here, I would take offence at your response.
    But the fact is, you know so little about what you are talking about it is not worth the time a logician to response to your elementary errors.
    While I applaud your efforts to understand the basics of logical arguments, I must tell you that your limited experience in this area can be a real determent to those who hope to learn from your posts. Actually your own confusion is a determent to the understanding you hope to contribute by your postings.
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  10. #10
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    Hello, olenka!

    I forgot the name of this rule: . \bigg[(p \vee q) \wedge (\sim p)\bigg] \to q
    . . I'll call it "Detchment", okay?


    How to prove that this below argument is valid,
    using established Rules of reasoning and Logical Equivalences
    (using vertical statement - reason format.)

    " alt="\bigg\{ [(\sim p \vee q) \to r] \wedge [r \to (s \vee t)] \wedge (\sim s \wedge \sim u) \wedge (\sim u \to \sim t)\bigg\} \:\to\ " />

    On the left side, we have:

    . . \bigg[(\sim p \vee q) \to r\bigg] \wedge \bigg[r \to (s \vee t)\bigg] \wedge (\sim s) \wedge (\sim u) \wedge \bigg[(\sim u) \to (\sim t)\bigg] \qquad\text{Given}

    . . \bigg[\sim(\sim p \;\vee\; q) \;\vee\; r\bigg] \wedge \bigg[\sim r \;\vee (s \;\vee \;t)\bigg] \wedge \;(\sim s)\; \wedge\; (\sim u) \wedge\; \bigg[u \vee \sim t\bigg] \qquad \text{Equiv. of impl'n}

    . . \bigg[(p \wedge \sim q) \vee r\bigg] \wedge \bigg[\sim r \vee(s \vee t)\bigg] \wedge (\sim s) \wedge (\sim u) \wedge \bigg[u \vee \sim t\bigg] \qquad \text{DeMorgan}

    . . \bigg[r \vee (p \wedge \sim q)\bigg] \wedge \bigg[(s \vee t) \vee \sim r \bigg] \wedge (\sim s) \wedge \underbrace{(\sim u) \wedge \bigg[u \vee \sim t\bigg]} \qquad \text{Commutative}
    . . \bigg[r \vee (p \wedge \sim q)\bigg] \wedge \bigg[(s \vee t) \vee \sim r\bigg] \wedge (\sim s)\quad \wedge\quad (\sim t)\qquad \text{Detachment}

    . . \bigg[r \vee (p \vee \sim q)\bigg] \wedge \underbrace{\bigg[\sim(\sim s \wedge \sim t) \vee \sim r\bigg] \wedge \bigg[\sim s \wedge \sim t\bigg]} \qquad\text{ DeMorgan}

    . . . . . . \underbrace{\bigg[r \vee (p\; \wedge \sim q)\bigg] \qquad\wedge \qquad \sim r} \qquad\qquad\text{ Detachment}

    . . . . . . . . . . . . .  p\; \wedge \sim q\qquad\qquad\text{ Detachment}


    And: . (p \:\wedge \sim q) \to p\quad\hdots . Forgot the name of this rule, too

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  11. #11
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    Wink

    Thanks to all of you for your effortt!!!!!!!
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