Results 1 to 3 of 3

Math Help - Laws of logic question

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517

    Laws of logic question

    hey guys, i have this question.

    (((p \rightarrow r ) \rightarrow p ) \wedge q ) \wedge (-((p \rightarrow r ) \rightarrow p ))

    my answer after simplifing boths sides i got

     (p \wedge q ) \wedge (-p)

    which then using associative law gives

     q \wedge ( p \wedge -p )

    then inverse law

     q \wedge T

    then Annihilation law

     T

    so this is a tautology. Can anyone confirm this is correct? thanks!

    EDIT: Sorry just noticed i posted this thread in the wrong section. Needs to be moved to discrete maths
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jvignacio
    Quote Originally Posted by jvignacio View Post
    hey guys, i have this question.

    (((p \rightarrow r ) \rightarrow p ) \wedge q ) \wedge (-((p \rightarrow r ) \rightarrow p ))

    my answer after simplifing boths sides i got

     (p \wedge q ) \wedge (-p)

    which then using associative law gives

     q \wedge ( p \wedge -p )

    then inverse law

     q \wedge T

    then Annihilation law

     T

    so this is a tautology. Can anyone confirm this is correct? thanks!

    EDIT: Sorry just noticed i posted this thread in the wrong section. Needs to be moved to discrete maths
    I assume that -p means \neg p (or \sim p) in which case your answer is not correct. p \land \neg p = F. So the result is F.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,663
    Thanks
    603
    Helolo, jvignacio!

    i have this question.

    \bigg( \left[(p \to r) \to p \right] \wedge q\bigg) \wedge \bigg(\sim [(p \to  r ) \to p)]\bigg)

    Let: . s \:=\:(p\to r)\to p


    Then we have: . (s \wedge q)\: \wedge \sim\! s

    . . . . . . . . . . =\;s \wedge (q \:\wedge \sim\! s). . Assoc. Property

    . . . . . . . . . . =\; s \wedge (\sim\!s \wedge q) . .Comm. Property

    . . . . . . . . . . =\; (s \:\wedge \sim\!s) \wedge q . . Assoc. Property

    . . . . . . . . . . =\; F \wedge \:q

    . . . . . . . . . . =\; F


    It is a Contradiction.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laws of logic
    Posted in the Discrete Math Forum
    Replies: 14
    Last Post: August 29th 2011, 02:30 AM
  2. Laws of logic question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: June 11th 2010, 12:22 AM
  3. Logic Laws Question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: April 9th 2010, 09:27 PM
  4. Help with Logic Laws
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: April 9th 2010, 07:22 PM
  5. laws of logic question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 14th 2010, 09:11 AM

Search Tags


/mathhelpforum @mathhelpforum