Results 1 to 5 of 5

Thread: Rewriting expressions with logic laws (2)

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

    Rewriting expressions with logic laws (2)

    Rewrite: $\displaystyle (p \wedge (q \to (p \vee (p \wedge q)))) \wedge q$ and determine if its tautology, contradiction or neither. Using truth table I got neither.

    My working out:

    - $\displaystyle (p \wedge (q \to (p \vee (p \wedge q)))) \wedge q$
    - $\displaystyle (p \wedge (q \to p)) \wedge q$
    - $\displaystyle (p \wedge (\sim q \vee p)) \wedge q$
    - $\displaystyle (p \wedge (p \vee \sim q)) \wedge q$
    - $\displaystyle ((p \wedge p) \vee \sim q) \wedge q$
    - $\displaystyle (p \vee \sim q) \wedge q$

    Now im a little stuck. After the truth table, im assuming I have to finish off with just a "p". Any help would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Hi

    Since your formula has the form $\displaystyle A\wedge q,$ finding only $\displaystyle p$ will be hard.

    In your other post, you used $\displaystyle p\wedge(p \vee\neg q)\equiv p.$ Applying this to your third or fourth line will allow you to conclude.

    (What you did for line 4 $\displaystyle \rightarrow$ line 5 is wrong)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by clic-clac View Post
    Hi

    Since your formula has the form $\displaystyle A\wedge q,$ finding only $\displaystyle p$ will be hard.

    In your other post, you used $\displaystyle p\wedge(p \vee\neg q)\equiv p.$ Applying this to your third or fourth line will allow you to conclude.

    (What you did for line 4 $\displaystyle \rightarrow$ line 5 is wrong)
    Thanks for your help, but I will only end up with$\displaystyle p \wedge q$ and not $\displaystyle p$. Is that still fine?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Yes, I suggest you redo your table.
    Note that even if $\displaystyle p$ is true, as soon as $\displaystyle q$ is false, the formula is false; thus the formula cannot be equivalent to $\displaystyle p$!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by clic-clac View Post
    Yes, I suggest you redo your table.
    Note that even if $\displaystyle p$ is true, as soon as $\displaystyle q$ is false, the formula is false; thus the formula cannot be equivalent to $\displaystyle p$!
    Are you referring to the truth table? To re-do it?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rewriting expressions with logic laws (4)
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Apr 13th 2010, 03:14 AM
  2. Rewriting expressions with logic laws (3)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Apr 13th 2010, 02:40 AM
  3. Rewriting expressions with logic laws (5)
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Apr 13th 2010, 02:25 AM
  4. Rewriting expressions with logic laws
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: Apr 10th 2010, 04:21 PM
  5. rewriting expressions
    Posted in the Algebra Forum
    Replies: 12
    Last Post: Sep 5th 2007, 05:37 PM

Search Tags


/mathhelpforum @mathhelpforum